College Physics, 4th Edition Knight, Jones, FieldKnight Solution Manual by TBtutors

Type:

Solution Manual

Resource:

College Physics

Edition:

4th Edition

Author(s):

Randal Knight Brian Jones Stuart Field

Instructor’s Solutions Manual for College Physics: A Strategic Approach Chris Porter

College Physics: A Strategic Approach Fourth Edition

Randall D. Knight Brian Jones Stuart Field

New York, NY

Director of Portfolio Management: Jeanne Zalesky Senior Portfolio Manager: Darien Estes Editorial Assistant: Kristen Stephens Content Producer: Martha Steele

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REPRESENTING MOTION

Q1.1. Reason: The softball player starts with an initial velocity but as he slides he moves slower and slower until coming to rest at the base. The distance he travels in successive times will become smaller and smaller until he comes to a stop. See the figure below.

Assess: Compare to Figure 1.10 in the text.

Q1.2. Reason:

Assess: The dots are equally spaced until the brakes are applied to the car. Equidistant dots indicate constant average speed. On braking, the dots get closer as the average speed decreases.

Q1.3. Reason:

© Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1-1

1-2

Chapter 1

Assess: The spacing between dots is initially large, since the initial speed with which the bush baby leaves the ground is large. As the bush baby rises and gravity slows the ascent, the speed decreases and therefore the spacing between adjacent dots in the motion diagram decreases.

Q1.4. Reason: As the ball drops from the tall building t02.he ball will go faster and faster the farther it falls under the pull of gravity. The motion diagram should show the displacements for later times to be getting larger and larger. The successive displacements in the diagram given in the text get smaller and smaller. So the diagram given in the problem is incorrect. The correct diagram is below.

Assess: Compare to Figure 1.5 in the text, which shows a motion diagram for two objects falling under the influence of gravity. The displacements increase during the fall of the object as we reasoned.

Q1.5. Reason: Position refers to a location in a coordinate frame. A displacement is the difference between two positions. In general, displacement is a vector and requires a direction. But in this one-dimensional case, we ignore that subtlety. The four miles Mark and Sofia walked definitely refer to a difference between their starting and ending positions. There is no information given about a reference frame. So it is more reasonable to associate the four miles with a displacement (magnitude) than with an absolute position. Assess: Mark and Sofia’s position could be specified as (for example) 100 m west of a particular intersection. But what is described is a difference between two positions. This is more like a displacement magnitude. Note that if their starting point had been the origin of a coordinate system, then both Mark and Sofia would be correct. Q1.6. Reason: The distance you travel will be recorded on the odometer. As you travel, the distance you travel accumulates, is recorded by the odometer, and is independent of the direction of travel. Your displacement is the difference between your final position and your initial position. If you travel around a 440 m track and end up where you started, you have traveled 440 m; however, since you ended up where you started, your change in position and hence displacement is zero. Assess: If you watch a track meet, you will observe the 440-m race. As you watch the race, it is obvious that the runners travel a distance of 440 m (assuming they complete the race). Yet since they end up where they started, their final position is the same as their initial position and hence their displacement is zero. Q1.7. Reason: Both speed and velocity are ratios with a time interval in the denominator, but speed is a scalar because it is the ratio of the scalar distance over the time interval while velocity is a vector because it is the ratio of the vector displacement over the time interval. Speed and velocity have the same SI units, but one must specify the direction when giving a velocity. An example of speed would be that your hair grows (the end of a strand of hair moves relative to your scalp) at a speed of about 0.75 in/month. An example of velocity (where direction matters) would be when you spring off a diving board. Your velocity could initially be 2.0 m/s up, while later it could be 2.0 m/s down.

Representing Motion

1-3

Assess: Saying that a velocity has both magnitude and direction does not mean that velocity is somehow “better” and that speeds are never useful. Sometimes the direction is unimportant and the concept of speed is useful. In other cases, the direction is important to the physics, and velocity should be cited. Each shows up in various physics equations.

Q1.8. Reason: Since the velocity of the skateboard is negative during the whole time of its motion, it is moving in the negative direction the entire 5 seconds. In order to move closer to the origin, which is in the positive direction relative to the starting point of the skateboard, the skateboard must have had a velocity in the positive direction for some time. Since the velocity is always negative, the skateboard must be farther from the origin than initially. Assess: Velocity gives direction of motion since it refers to displacements and not only distance traveled. If velocity is always negative, displacement will be negative also. Q1.9. Reason: If the position of the bicycle is negative it is to your left. The bicycle’s velocity is positive, or to the right, so the bicycle is getting closer to you. Assess: If the initial position had been positive and the velocity positive, the bicycle would be getting farther away from you. Q1.10. Reason: Since the jogger is running around a track, she returns to her starting point at the end of the lap. Since her final position is the same as her initial position, her displacement is 0 m. Velocity is defined as velocity =

displacement in a given time interval time interval

So her average velocity is 0 m/s! However, though her displacement is 0 m, the actual distance she traveled is 400 m. Her average speed is not zero, since speed is defined in terms of distance, not displacement. speed =

distance traveled in a given time interval time interval

400 m

Her average speed is then 100 s = 4 m/s The second friend is correct. Since the motion of a runner is not uniform we can only calculate average velocity and average speed. Assess: This problem illustrates a very important difference between speed and velocity. Speed depends on total distance traveled. Velocity depends on displacement, which only takes into account the starting and ending points of a motion.

Q1.11. Reason:

Assess: The dots get farther apart and the velocity arrows get longer as she speeds up.

Q1.12. Reason: The child will be traveling with a constant velocity until hitting the rocky patch. During the constant velocity part of the motion the motion diagram should show equal displacements during each time interval. After hitting the rocky patch, the sled will start slowing. After this point the motion diagram should show everdecreasing displacements and ever-decreasing velocity vectors. The motion diagram is below.

Assess: Compare to Figure 1.11 in the text where a similar motion is illustrated. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1-4

Chapter 1

Q1.13. Reason: The initial velocity is zero. The velocity increases and the space between position markers increases until the chute is deployed. Once the chute is deployed, the velocity decreases and the spacing between the position markers decreases until a constant velocity is obtained. Once a constant velocity is obtained, the position markers are evenly spaced. See the following figure.

Assess: Knowing the velocity of the jumper will increase until the chute is deployed and then rapidly decrease until a constant descent velocity is obtained allows one to conclude that the figure is correct.

Q1.14. Reason: The tennis ball falls freely for the three stories under the pull of gravity. Since gravity is pulling it downward, its speed increases with time. It strikes the ground and very quickly slows down to a stop (while compressing the ball) then bounces back upward (while the ball decompresses). After the bounce, it travels upward while still under the pull of gravity. As it is traveling upward the pull of gravity decreases its velocity to zero at a height of two stories. The downward and upward motions of the ball are shown in the figure below. The increasing length of the arrows during the downward motion indicates increasing velocity. The decreasing length of the arrows during the upward motion indicates the particle slowing down.

Representing Motion

1-5

Assess: Compare Figure 1.5 in the text for the downward motion. Note that gravity always pulls downward, no matter what the direction of motion of something under its influence. So the ball will constantly be slowing down during its upward motion.

Q1.15. Reason:

Assess: The car (particle) moves at a constant speed v so the distance between the dots is constant. While turning v remains constant, but the direction of v changes.

Q1.16. Reason: No. The 4 m vector is longer than the other two combined. The 1 m and 2 m vector cannot undo the 4 m vector. Assess: This is easy to see if we think of these vectors as displacement vectors. If a person moves 4 m in a straight line, it is obvious that they cannot return to their starting point by moving only 1 m and then 2 m in any direction. Q1.17. Reason: We do not believe Travis. 55 m is more than 165 ft. It is not reasonable to think that a unicycle could move that quickly. Assess: The evaluation may be easier in different units. Consider converting to miles per hour: 55 m 1 mi 3600 s   = 120 mph s 1610 m 1 hr

This is twice the speed limit on many freeways, and not reasonable for a unicycle.

Q1.18. Reason: The student changes direction during the walk. Her first displacement vector is 1.0 m west, and her second is in a different direction, 1.0 mi north. We need to use the rules for addition of vectors to find the distance she is from her starting point. The distance is the length of her net displacement vector. See the diagram below.

Her displacement is Dnet = D1 + D2  The length of Dnet is given by the Pythagorean theorem since the displacement vectors form a right triangle. Her net displacement vector is along the hypotenuse of the triangle. 2

AC = AB + BC = (10 mi)2 + (10 mi)2 = 14 mi

1-6

Chapter 1

We can find her direction also, though this is not asked for in this question. The angle would be given by

 1.0 mi   = arctan(1.0) = 45  1.0 mi 

 = arctan 

So the student’s displacement is 1.4 mi 45 north of west. The correct answer is B. Assess: To find a net displacement the laws of vector addition must be used. The net displacement here is the vector addition of the displacements that made up the motion.

Q1.19. Reason: Since the rock is above the origin the position is positive; since it is still moving upward the velocity is also positive. Hence, the correct answer is A. Assess: After it gets to the top and starts back down, the position will still be positive, but the velocity will be negative. Q1.20. Reason: Because of the numbering of the dots, we see the object is moving to the left. It is slowing because the dots are getting closer together. The choice that fits this scenario is a cyclist moving to the left and braking to a stop. So choice B is correct. Assess: If the dots were numbered in reverse order then choice C would be correct. Q1.21. Reason: Because the dots are getting farther apart to the right (and the numbers are increasing to the right) we know that the object is speeding up. The choice that best fits that is a car pulling away (to the right) from a stop sign. So the correct choice is C. Assess: An ice skater gliding (choice A) would likely have nearly constant velocity (constant spacing between dots). The motion diagram for a plane braking (choice B) might look like the given diagram with the dots numbered in reverse order. The pool ball reversing direction (choice D) would have dot numbers increasing in one direction at first but then going the other way. Q1.22. Reason: The bird changes direction during the flight. The first displacement vector is 3.0 km due west, and her second is in a different direction, 2.0 km due north. We can use the rules for addition of vectors to find the distance the bird is from her starting point. The distance is the length of her net displacement vector. See the diagram below.

The bird’s net displacement is Dnet = D1 + D2  The length of Dnet is given by the Pythagorean theorem since the displacement vectors form a right triangle. The net displacement vector is along the hypotenuse of the triangle. 2

AC = AB + BC = (30 km)2 + (20 km)2 = 36 km We can find the direction of the displacement also, though this is not asked for in this problem. In general, to specify a displacement a length and a direction must be specified. The angle in the diagram would be given by  2.0km   = 34  3.0km 

 = arctan 

So the bird’s displacement is 3.6 km 34 north of west. The correct answer is C. Assess: The net displacement is made up of the vector addition of the displacements that made up the motion.

Representing Motion

1-7

Q1.23. Reason: The speed is the distance divided by the time. speed =

distance 030 km  1000 m  1 min  =    = 10 m/s time 50 min  1 km  60 s 

So the correct choice is B. Assess: 1 m/s does seem like a reasonable speed for a seal in water.

Q1.24. Reason:

()

1 = 3 and either of them can be deduced from tan −1 23 = 337

2 = 90 − 2(1) = 90 − 2(337) = 23 So the correct choice is A. Assess: By a quick back-of-the-envelope sketch we can see that choices C and D are much too large.

Q1.25. Reason: This is a simple unit conversion problem: 400 m 1 mi 3600 s   = 17.2 mph 51.9 s 1610 m 1 hr

So the correct choice is C. Assess: This is the speed of a slow-moving car, and is reasonable for a very fast-moving human.

Q1.26. Reason: The second rule on significant figures says that when adding two numbers the number of decimal places in the answer should match the smallest number of decimal places of any number used in the calculation. There are three decimal places in 0.532 m, but only two in 3.24 m, so our answer must have only two decimal places. Therefore the correct choice is B: 3.77 m. Assess: On contemplation we realize that we don’t know the 3.24 m number better than to the nearest cm, and that will also be true of the answer, even though we may know the other number to the nearest millimeter. Q1.27. Reason: When multiplying numbers, the correct number of significant digits is the smaller of the numbers of significant digits in the two numbers. That is, here we have four significant figures in 109.7 m and three in 48.8 m, and so our product must have three significant digits:

(109.7 m )  ( 48.8 m ) = 5.35 103 m2 The correct answer is B. Assess: Our final answer has three digits of precision, since we multiplied by something with only three digits of precision. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1-8

Chapter 1

Q1.28. Reason: This is a straightforward unit conversion question.  36525 d  24 h  60 min  60 s  17 457  109 years = 457  109 yr      = 144  10 s 1 yr 1 d 1 h 1 min     

The correct choice is D. Assess: Notice that even though we have more significant figures in some of the conversion factors (we accounted for leap year, and the other factors are exact and have as many significant figures as we need) that we obey the significant figure rules and report the answer to the same number of significant figures as the number of significant figures in the least precisely known number in the calculation (the age in years).

Q1.29. Reason: We are given an equation for density and are asked to calculate the density of the earth given its mass and volume. However, the units must be converted before the calculation is done since we’re given volume in km3 and the answer must be given in terms of m3.  103 m  103 m  103 m  V = (108  1012 km3 )   1 km   1 km   1 km      3

 103 m   109 m3  = (108  1012 km3 )  = (108  1012 km3 )   1 km   1 km3      = 108  1021 m3

Note carefully that we needed three conversion factors for the conversion from km to m here since we are dealing with cubic kilometers. Three factors are needed to cancel the factor of km3 = km  km  km. So the density is

=

M 594  1024 kg = = 550  103 kg/m3 V 108  1021 m3

The correct choice is A. Assess: For cubic and square units (or units to any power) you must include the correct number of conversion factors to convert every factor in the original quantity. Since the density of water is 10 103 kg/m3 , it seems reasonable that the earth would be 5.5 times as dense.

Problems P1.1. Strategize: A motion diagram consists of images or positions of an object shown with equal time intervals between each successive image. This is similar to images taken by a camera with fixed timing between images. Prepare: Frames of the video are taken at equal intervals of time. As a result, we have a record of the position of the car at successive time equal intervals – this information allows us to construct a motion diagram. Solve:

Assess: Once the brakes are applied, the car slows down and travels a smaller distance during each successive time interval until it stops. This is what the car in the figure is doing.

Representing Motion

1-9

P1.2. Strategize: For a motion diagram we want to make dots indicating the person’s position, one for each fixed time interval. In this case, the natural time interval to use is one minute. Prepare: We will assume the man travels with constant speed while he’s walking and riding. There are three parts to motion, each with constant speed. Solve: The diagram is below.

Assess: While the man is walking his speed is less than when riding the bicycle, so the dots should be more closely spaced during that part of the trip.

P1.3. Strategize: To draw an accurate motion diagram, we must consider when Amanda speeds up and when she slows down, and we must space the dots in our diagram accordingly. Prepare: As the elevator begins to rise, its speed changes from zero to some other speed. An elevator typically maintains a steady speed for a while, and then slows as it reaches the desired floor. So our diagram should start with very small but increasing spacing between dots (getting started), then have wider spaced dots with constant spacing for some time (constant speed), and finally dots with decreasing spacing (slowing to a stop). Solve:

Assess: The speed (and therefore the spacing between dots) increases from 1 to 3, then stays constant from 3 to 5, and decreases from 5 to 7.

P1.4. Strategize: We are asked to find a displacement, which does not depend on the origin. Therefore, we expect to find the same answer to parts (a) and (b). Prepare: We must determine Sue’s position, being careful of the sign of the displacement. Solve: (a) Refer to the figure below.

If Sue’s home is the origin of the x-axis, she is 2 mi to the right of the origin. This is the positive side of the axis, so her position x = +2 mi. (b) Refer to the figure below.

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Chapter 1

Taking the cinema as the origin of the x − axis, then Sue is 3 mi to the left of the origin. This is the negative portion of the axis, so Sue’s position is x = −3 mi Assess: Displacement along a straight line is a signed quantity. It’s important to indicate the sign when reporting a displacement relative to a chosen origin. See Figures 1.13 and 1.12 in the text for examples of both negative and positive displacements, respectively.

P1.5. Strategize: The displacement is the difference between two positions. It does not depend on the origin. Prepare: To find Sue’s displacement, we want to find the difference between her initial position between her home and the cinema, and her position at her home. Solve: It is graphically clear that x = xf − xi = 2 mi. This is the answer for both parts (a) and (b). Assess: Sue’s position depends on where the origin is. But the displacement is the difference between two points (initial and final), and that is independent of what point in space we label as our origin.

P1.6. Strategize: The displacement is the difference between the initial and final positions. Prepare: In this case, we know that the initial position is at the 65 mm mark and the final position is at the 42 mm mark. We find the displacement using x = xf − xi . Solve: We have x = xf − xi = ( 42 mm ) − ( 65 mm ) = −23 mm.

Assess: This is a reasonable distance for a paramecium under a microscope, and since the final position is a smaller number than the initial number, the negative sign of the displacement also makes sense.

P1.7. Strategize: Displacement is the difference between a final position xf and an initial position xi . Prepare: The displacement can be written as x = xf − xi , and we are given that xi = 23 m and that  x = −45m Solve:  x = xf − xi Since we want to know the final position we solve this for xf 

xf = xi + x = 23 m + (−45 m) = −22 m Assess: A negative displacement means a movement to the left, and Keira has moved left from x = 23 m to x = −22 m.

P1.8. Strategize: Displacement along a straight line is a signed quantity, and is given by x = xf − xi . Prepare: Since this problem asks for displacements relative to Mulberry Road, we will choose Mulberry Road as the origin. Solve: Since Mulberry Road was chosen as the origin and x increases to the east, the final position of the car is xf = −14 mi. We are told that the displacement of the car was x = −23 mi, so during its motion the car traveled 23 mi west. We can use the definition of displacement, x = xf − xi , to find the car’s initial position.

xi = xf −  x = (−14 mi) − (−23 mi) = + 9 mi The car started 9 mi east of Mulberry Road.

Representing Motion

1-11

Assess: Note the choice of origin for measuring positions and displacements is arbitrary, but in many cases it’s convenient to choose a reference point in the problem as the origin. Here, all displacements and positions were mentioned relative to Mulberry Road, which was chosen as the origin and made the calculations direct.

P1.9. Strategize: We must combine three different displacements to determine the total displacement. Prepare: We have been given three different displacements. The problem is straightforward since all the displacements are along a straight east-west line. All we have to do is add the displacements and see where we end up. Solve: The first displacement is x1 = 500 m east, the second is x2 = 400 m west and the third displacement is x3 = 700 m east. These three displacements are added in the figure below.

From the figure, note that the result of the sum of the three displacements puts the bee 800 m east of its starting point. Assess: Knowing what a displacement is and how to add displacements, we are able to obtain the final position of the bee. Since the bee moved 1200 m to the east and 400 m to the west, it is reasonable that it would end up 800 m to the east of the starting point.

P1.10. Strategize: We are told the guard has a steady pace, meaning there is only one speed to calculate (the average speed). This speed is defined in Equation 1.1 of the text. Prepare: Speed is the distance traveled in some time interval divided by the length of the time interval (at least in this case of a steady pace, in which average speed and speed refer to the same thing). Solve: From Equation 1.1, speed =

distance traveled in a given time interval 110 m = = 046 m/s time interval 240 s

Assess: Someone walking at a brisk pace will easily travel more than 1 m/s. However, since a guard would travel at more like a stroll, this is a reasonable speed.

P1.11. Strategize: In all cases, objects are moving at a steady pace. So speed is just the distance divided by the time. Prepare: We are asked to rank in order three different speeds, so we simply compute each one according to Equation 1.1: speed = Solve: (i) Toy (ii) Ball (iii) Bicycle

distance traveled in a given time interval time interval 015 m = 0060 m/s 25 s 23 m = 42 m/s 055 s 060 m = 80 m/s 0075 s

80 m = 40 m/s 20 s So the order from fastest to slowest is bicycle, ball, cat, and toy car. (iv) Cat

1-12

Chapter 1

Assess: We reported all answers to two significant figures as we should according to the significant figure rules. The result is probably what we would have guessed before solving the problem, although the cat and ball are close. These numbers all seem reasonable for the respective objects.

P1.12. Strategize: In the motion diagram, we are told the time interval between each pair of dots. The positions have been labeled. Thus we have sufficient information to determine speeds on different time intervals. Prepare: We can use Equation 1.2 to calculate the horse’s velocity at the different times. Solve: Since the dots are spaced at equal intervals of time, and there is one dot between the time 10 s and 30 s, the spacing between the dots indicate a 10 s time interval. The dot between 10 s and 30 s will mark a time of 20 s. The horse is moving to the left, as time increases to the left, so the rightmost dot must be at 0 s. We will use the definition of velocity in Equation 1.2, v = x/t Looking at Figure P1.12, it should be noted that you can’t determine the distance information to better than 10 m and certainly not to 1 m. As a result 100 m has two significant figures (you know it is between 90 m and 110 m) and 50 m has one significant figure (you know it is between 40 m and 60 m). For this reason the answer to part (b) should have only one significant figure. (a) Referring to the P1.12 in the text, xf = 500 m, xi = 600 m, tf = 10 s, ti = 0 s, so v=

x 500 m − 600 m −100 m = = = −10 m/s t 10 s − 0 s 10 s

(b) Here, xf = 300 m, xi = 350 m, tf = 40 s, ti = 30 s, so

x 300 m − 350 m −50 m = = = −5 m/s t 40 s − 30 s 10 s

x 50 m − 250 m −200 m = = = −10 m/s t 70 s − 50 s 20 s

Assess: Displacement and velocities are signed quantities. Since the x-axis increases to the right and the horse is traveling to the left, we should expect all the velocities to be negative.

P1.13. Strategize: Average velocity is defined as the displacement x divided by the time interval t Prepare: We are given t = 35 s, but we will do a preliminary calculation to find the displacement.

x = xf − xi = −47 m − (−12 m) = −35 m Solve:

x −35 m = = −10 m/s t 35 s

Assess: The answer is reasonable, and agrees with the approximate walking speed estimated in Example 1.5. The negative sign tells us that Harry is walking to the left.

P1.14. Strategize: Average velocity is defined as the displacement x divided by the time interval t Prepare: We are given the initial and final positions, and the time interval, so we can use the definition of velocity, which is given in Equation 1.2. Solve: xf = 3 m, xi = −12 m, t = 10 s, so v=

x 3 m − (−12) m 15 m = = = +15 m/s t 10 s 10 s

which should be reported as +2 m/s to one significant figure. Assess: Note that it’s important to keep track of signs on positions and displacements in equations.

Representing Motion

1-13

x . t Prepare: Since we are told the times after intervals of 100 m, the x is the same over each interval. Only t changes. We need to find the shortest time interval, since that will correspond to the highest speed. Solve: We find the duration of each of the four intervals:

P1.15. Strategize: We want to find the highest speed, meaning the highest value of: speed =

t1 = 11.20 s

t2 = ( 21.32 s ) − (11.20 s ) = 10.12 s t3 = ( 31.76 s ) − ( 21.32 s ) = 10.44 s t4 = ( 43.18 s ) − ( 31.76 s ) = 11.42 s

(a) Clearly the second 100 m was done in the shortest amount of time. So the second 100 m was the fastest. x 100 m = = 9.88 m/s (b) speed = t 10.12 s Assess: This is about 22 mph: extremely fast for a human, but still entirely plausible.

P1.16. Strategize: These are all simple unit conversions. SI units refer to the base units such as meters and seconds, with no scaling prefix. For example, we want to write 1 s as 10−6 s. We will keep all original significant digits, since these metric conversions are exact. Prepare: We first collect the necessary conversion factors: 1 s = 10−6 s; 1 km = 103 m; 1 m = 102 cm; 1 h = 60 min; 1 min = 60 s. Solve:  10−6 s  = 912  10−6 s (a) 912 s = (912 s)   1 s     103 m  = 342  103 m (b) 342 km = (342 km)   1 km    −2  cm   10 m  1 ms  2 (c) 44 cm/ms = 44    = 44  10 m/s    ms   1 cm   10−3 s  3  1h  km   10 m  (d) 80 km/h = 80    = 22 m/s   2    h   1 km   36  10 s 

Assess: The conversion factors are applied in such a manner that we obtain the desired units. Scientific notation is used and the answer has no more significant figures than the starting number.

P1.17. Strategize: These are all simple unit conversions. SI units refer to the base units such as meters and seconds, with no scaling prefix. In particular, we do not want to use imperial units like feet or inches. Prepare: We first collect the necessary conversion factors: 1 in = 2.54 cm; 1 cm = 10−2 m; 1 ft = 12 in; 39.37 in = 1 m; 1 mi = 1.609 km; 1 km = 103 m; 1 h = 3600 s. Solve:  254 cm   10−2 m  (a) 80 in = 80 (in)   = 020 m   1 in   1 cm 

 ft   12 in  1 m  (b) 66 ft/s = 66      = 20 m/s  s   1 ft  3937 in  3  mi   1609 km   10 m   1 h  (c) 60 mph = 60        = 27 m/s  h   1 mi   1 km   3600 s 

1-14

Chapter 1

P1.18. Strategize: The SI unit of time is seconds. So this is a simple time conversion from various units into seconds. Prepare: We first need the necessary conversion factors: 1 year = 365.25 days; 1 day = 24 hours; 1 h = 3600 s.  3600 s  3 Solve: (a) 10 hour = 10 (h)   = 3600 s = 36 10 s 1 h  

 24 hours  3600 s  4 (b) 10 day = 10 (d)    = 86  10 s  1 d  1 h   36525 days   864  104 s  7 (c) 10 year = 10 (year)   = 32  10 s   1 year 1 d   

Assess: Since the given information contains only two significant figures, all answers contain only two significant figures.

P1.19. Strategize: Review the rules for significant figures in Section 1.4 of the text. Prepare: Pay particular attention to any zeros and whether or not they are significant. Note for example that the leftmost zero in (c) is a place-holder and is not a significant digit, but the right-most zero in (c) specifies an additional digit of precision and is significant. Solve: (a) The number 6.21 has three significant figures. (b) The number 62.1 has three significant figures. (c) The number 0.620 has three significant figures. (d) The number 0.062 has two significant figures. Assess: In part (c), the final zero is significant because it is expressed and in part (d), the second zero locates the decimal point but is not significant. P1.20. Strategize: Review the rules for significant figures in Section 1.4 of the text. Prepare: Pay particular attention to any zeros and whether or not they are significant. Solve: (a) The number 0.621 has three significant figures. (b) The number 0.006200 has four significant figures. (c) The number 1.0621 has five significant figures. (d) The number 6.21103 has three significant figures. Assess: In part (b), the initial two zeros place the decimal point. The last two zeros do not have to be there, but when they are they are significant. P1.21. Strategize: Review the rules for significant figures in Section 1.4 of the text. Prepare: Pay particular attention to the rules for addition (and subtraction) and multiplication (and division). Solve: (a) 33.3 25.4 = 846 (b) 33.3 − 25.4 = 7.9 (c) 333 = 577 (d) 333.3  25.4 = 13.1 Assess: In part (a) the two numbers multiplied each have three significant figures and the answer has three significant figures. In part (b), even though each number has three significant figures, no information is significant past the tenths column. As a result, the answer is expressed only to the tenths column. In part (c), the number and the answer both have three significant figures. In part (d) the answer is expressed to three significant figures since this is the least number of significant figures in either of the two numbers in the problem.

P1.22. Strategize: The number of significant digits should reflect the certainty of the measurement. A digit that is not certain should not be specified. Prepare: The uncertainty is nearly a half an inch in either direction; this is about one centimeter in either direction, so we’ll express the answer to the nearest centimeter. Solve: Convert the man’s height to inches: 6 ft 1in = 73 in  254 cm  1 m  73 in    = 185 m  1 in  100 cm  Assess: To round this to two significant figures, 1.9 m, would give the wrong impression of the man’s height. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Representing Motion

1-15

P1.23. Strategize: This is a simple unit conversion. Prepare: To convert, we will use the fact that each foot contains exactly 12 inches, and each inch is 2.54 cm, or 0.0254 m.  12 in  254 cm  1 m  3 Solve: We have: 29,029 ft     = 8.8480  10 m.  1 ft  1 in  100 cm  Assess: Note that we have used exact conversions, such that there was no loss of significant digits in the conversion. The information was given to five significant digits, and we still have five significant digits. P1.24. Strategize: This is an estimation problem. We will need to assemble estimates from life experience, and many answers are possible. Prepare: First make an estimate of how much you cut and how often you mow. Let’s say you cut an inch per week. Then it’s just a matter of unit conversion to get it into m/s.  1 in   1 week   1 day  1 hr  254 cm  1 m  −8 Solve: v =        = 4  10 m/s 3  1 week   7 day   24 hr  33  10 s  1 in  100 cm  Assess: Since a week is very large compared to a second and a m is about 40 times the size of an inch, we are expecting a very small number. Even professional grass growing watchers don’t expect to see much activity in one second.

P1.25. Strategize: This is an estimation problem. We will need to assemble estimates from life experience, and many answers are possible. Prepare: My barber trims about an inch of hair when I visit him every month for a haircut. The rate of hair growth thus is one inch per month. We also need the conversions 1 in = 2.54 cm, 1 day = 24 h, 1 month = 30 days, 1 h = 3600 s, 1 cm = 10−2 m.

Solve: The rate of hair growth is −2  1 in   254 cm   10 m   1 month  1 d  1 h  −9       = 98  10 m/s  35  m/h    month   1 in   1 cm   30 d  24 h  3600 s 

Assess: Since we expect an extremely small number for the rate at which our hair grows per second, this figure is not unreasonable.

P1.26. Strategize: We are asked for the straight-line distance between two places. This is the same as the magnitude of the displacement between two locations, so we can view this as a displacement problem. Prepare: To find the total distance between the workplaces, we can calculate the displacement from Robin’s workplace to Carol’s workplace. Solve: The displacement from the house to Robin’s workplace is 7.5 km west. The displacement from the house to Carol’s workplace is 2.0 km north. Refer to the diagram below.

The total displacement between Robin’s workplace and Carol’s workplace is the length of the hypotenuse of the triangle in the figure. This is

D = ( DR )2 + ( DC )2 = (75 km)2 + (20 km)2 = 78 km Assess: Since the displacements are in different directions, we must treat them as vectors.

1-16

Chapter 1

P1.27. Strategize: We are asked for the straight-line distance between two places. This is the same as the magnitude of the displacement between two locations, so this could be viewed as a displacement problem. Prepare: In this problem we need to find the distance between Fort Collins and Greeley. We’ll use the Pythagorean theorem. Solve:

(18 km)2 + (31 km)2 = 36 km Assess: This is a reasonable distance between cities.

P1.28. Strategize: We are asked for the straight-line distance between two places. This is the same as the magnitude of the displacement between two locations, so this could be viewed as a displacement problem. Prepare: Joe and Max went at right angles to each other, so we can use the Pythagorean theorem to compute the distance between them. Solve: d = (325 km)2 + (055 km)2 = 33 km Assess: It is right that the difference between the two is greater than the distance either of them went; the answer seems about right for a skinny triangle.

P1.29. Strategize: This problem is asking for the straight-line distance between initial and final points, given a number of smaller legs of the journey. A picture may be beneficial. Prepare: We begin by sketching a drawing of the setup.

Clearly Veronica travels four blocks north, but then comes back one block south, such that she is finally three blocks north of her starting point. She also moves two blocks east. We can use the Pythagorean Theorem to combine these displacements in the east and north directions into one straight line distance. Solve: The component of displacement eastward is x = 2 ( 400 ft ) = 800 ft, and the displacement northward is y = 3( 280 ft ) = 840 ft. Then the total displacement is d = x2 + y 2 =

(800 ft )2 + (840 ft )2 = 1.16 103 ft.

Assess: This is a reasonable distance. One way of checking this would be to compare it to the distance required to walk two blocks east and then three blocks north, which is 1,640 ft. The straight-line distance is shorter, as it should be.

P1.30. Strategize: This is a displacement problem. Displacement is the difference between initial and final positions. Prepare: The displacement is the hypotenuse of a right triangle whose legs are 6 m (half the width of the garden) and 8 m, directed from xi at the top of the tree to xf at the top of the flower. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Representing Motion

1-17

(6 m)2 + (8 m)2 = 10 m Assess: Displacement is a vector and does have direction (as mentioned in the Prepare step), but we were only asked for the magnitude of the displacement. Solve: The length of the hypotenuse is

P1.31. Strategize: This is a displacement problem. The displacement is the difference between initial and final locations: x = xf − xi . Because this is two-dimensional motion, the vector nature of the displacement is important; we must provide a magnitude and a direction. Prepare: John’s displacement is the vector from his starting point to his ending point. Solve: Refer to the diagram below.

John stops at the southernmost end of the circle. His final position is 50 m west and 50 m south of his starting position since the radius of the circle is 50 m. We must find the displacement vector from the initial point to the final point. Point B is at the center of the circle. The displacement vector has a length of 2

AC = AB + BC = (50 m)2 + (50 m)2 = 71 m The angle in the diagram is  1.0m   = 45  1.0m 

 = arctan 

So the answer in (magnitude, direction) notation would be (71 m, 45 south of west).

1-18

Chapter 1

Assess: Compare the solution to the solution for Problem 1.27. Here it would be difficult to sum his displacement vectors along the circle, but this is not necessary since the displacement vector is always the vector from the initial position to the final position.

P1.32. Strategize: This problem requires us to use the concept of displacement (the difference between two positions) and relate it to the geometry of a lake. Prepare: If we look at the figure below, it is clear that the straight-line displacement is related to the radius of the circle in a simple way: ( x ) = R2 + R2 = 2R2 . 2

Solve: Using the geometric information from the figure above, we have x = 2 R, and we also know that

x = xf − xi = 180 m. Combining these two, we have R = (180 m ) / 2  D = 2 R = 2 (180 m ) = 255 m. Assess: Clearly this diameter is the right order of magnitude. We also expect that the straight line distance must be shorter than 2R (as seen from the figure), and it is.

P1.33. Strategize: We must consider two separate displacements and use them to find a total displacement. Prepare: Knowing that the total trip consists of two displacements, we can add the two displacements to determine the total displacement and hence the distance of the goose from its original position. A quick sketch will help you visualize the two displacements and the total displacement. Solve: The distance of the goose from its original position is the magnitude of the total displacement vector. This is determined as follows: d = (32 km)2 + (20 km)2 = 38 km Assess: A quick look at your sketch shows that the total distance should be larger than the largest leg of the trip and this is the case.

P1.34. Strategize: This problem contrasts the concepts of distance and displacement. Distance is what is measured by the odometer of a car, and depends on the path taken between two points. Displacement is the difference between two positions, and only depends on the two endpoints. Displacement does not depend on the path taken between the two endpoints. C 32 km = = 0509 km Prepare: We can solve for the radius of the track from C = 2 r  r = 2 2

Representing Motion

1-19

Solve:

a. The car has traveled half of the circumference, or 3.2 km/2 = 1.6 km b. The displacement of the car is the diameter (twice the radius) of the track north of its starting point.

D = 2(0509 km) = 10 km Assess: The distance traveled is longer than the displacement, as we would expect.

P1.35. Strategize: This problem involves motion in two orthogonal directions: vertical and horizontal. We can use trigonometry to relate the distances and the given angle. Prepare: It is helpful to draw a diagram.

Solve: Use knowledge about right triangles from trigonometry. tan 35 =

h  h = (100 m)(tan 35) = 61 m 100 m

The vulture loses 6.1 m of height as it flies 100 m horizontally. Assess: This small loss of height would indeed allow a vulture to glide long distances.

P1.36. Strategize: This problem involves a displacement in both the horizontal and vertical directions ( x and y ). Displacement is the difference between the final and initial locations, and does not depend on the twists and turns in between. Prepare: We calculate the x and y components separately and combine them using the Pythagorean Theorem. We associate the positive x direction with motion horizontally to the right in the map, and y will be the direction vertically upward on the map. We can determine the direction of the displacement using trigonometry. For example, we can determine the angle away from the positive x axis by using tan ( ) = y / x. Solve: The horizontal displacement is given by x = ( 0.80 mi ) + (1.20 mi ) cos ( 30) = 1.84 mi. The vertical displacement is given by y = ( −0.80 mi ) + (1.20 mi ) sin ( 30 ) = −0.20 mi. We combine these using the Pythagorean Theorem: d =

( x )2 + ( y )2 = (1.84 mi )2 + ( −0.20 mi )2 = 1.9 mi.

Then tan ( ) =

y −0.20 mi = = −6.2, meaning the displacement is 1.9 miles at an angle of 6.2 below the positive x 1.84 mi

x axis.

1-20

Chapter 1

Assess: This is a reasonable displacement, given that the sum of the legs of the trip is 2.8 miles. We expect the straight-line distance between initial and final points to be less than the sum of distances (with twists and turns) that Olivia took. And we see this is the case.

P1.37. Strategize: We must use simple trigonometry to relate the height and distance along the slope of the Great Pyramid. Prepare: We begin by making a sketch of the geometry described:

Clearly, we can relate the distance along the sloped side to the height by using the sine trigonometric function. h 139 m opposite h Solve: We know sin ( ) = = = 177 m. = , so D = sin ( ) sin ( 51.8) adjacent D Assess: We know that the distance along the slope should be greater than the height, and we know from the given angle that it should not very much greater. Our answer of 177 m makes sense.

P1.38. Strategize: This problem involves displacement vertically and horizontally. The distances along different directions can be related to the given angle using trigonometry. Prepare: It is helpful to draw a diagram.

Solve: Use knowledge about right triangles from trigonometry. h = (1500 m)(sin 10) = 260 m The hiker gains 260 m of elevation. Assess: 260 m is quite a bit less than 1500 m, but this is expected.

P1.39. Strategize: In this problem we have three displacements to add using the laws of vector addition. Prepare: We will use the diagram below.

Solve: A convenient place to place the origin is the origin of the motion. We could add the first two vectors and then add the third vector on to that result. However, by looking at the diagram, the total displacement in the x direction is 60 cm + 80 cm = 140 cm to the right. The total displacement in the y direction is 40 cm downward. The net displacement will reflect these two displacements, so the result of the vector addition will be a vector pointing 140 cm to the right and 40 cm downward. Considering the right triangle ABC in the diagram, the magnitude of the displacement vector is then 2

AC = AB + BC = (140 cm)2 + (40 cm)2 = 150 cm © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Representing Motion

1-21

Where we have assumed that the measurements in the problem have been given to two significant figures.

(

40 cm

)

The angle that the vector makes is  = arctan 140 cm = 16 below the positive x-axis. Assess: When adding any number of displacements, the net displacement is always the vector between the initial and final point of the motion.

P1.40. Strategize: We must determine the displacement of the ball using the information given about the motion down the field and toward the sidelines (orthogonal directions). Prepare: We know the initial position of the ball (the middle of the field). We know the ball travels 43 yards down field (the first displacement) and 26.5 yards to the side of the field (the second displacement). We can add these two displacements to obtain the net displacement (magnitude and direction). A quick sketch will help you visualize the two displacements and the net displacement.

Solve: The magnitude of the net displacement may be obtained by

d = (43 yd)2 + (265 yd)2 = 50 yd The direction of the net displacement may be obtained by  265 yd   = 32  43 yd 

 = tan −1 

The net displacement is 50 yd in a direction 32 degrees from straight down the middle of the field. Assess: Knowing the two displacements, we can add them to determine the net displacement (magnitude and direction). Looking at your sketch, you should expect the magnitude of the net displacement to be larger than either of the two displacements which contribute to the net displacement, and this is the case. Looking at your sketch, you should also expect the angle of the net displacement to be less than 45°, and this is the case.

P1.41. Strategize: Because this problem involves motion in two orthogonal directions, it will involve a right triangle. We can use trigonometry to relate given distances and angles. Prepare: Draw a diagram of the situation. Note the right triangle.

1-22

Chapter 1

Solve: Use knowledge about right triangles from trigonometry. (a) The length of the path is =

3 5 m = 46 m cos 40

(b) The height h above the ground is the vertical side of the triangle subtracted from 9.0 m.

h = 90 m − (35 m)(tan 40) = 61 m Assess: Because of the angle we expect the squirrel to lose less height than the horizontal distance traveled.

P1.42. Strategize: Because this problem involves motion in two orthogonal directions, it will involve a right triangle. We can relate the given distances to the required angle using trigonometry. Prepare: Draw a diagram of the situation. Note the right triangle.

Solve: Use knowledge about right triangles from trigonometry. (a) The angle of the path below the horizontal is

 8m   = 27  16 m 

 = arctan  (b) The distance d covered is

d = (16 m)2 + (80 m)2 = 18 m Assess: Because the squirrel descended less than the horizontal distance we expect the angle to be less than 45 degrees.

P1.43. We interpret the described motion in a motion diagram. We assume equal time intervals between dots, and we draw velocity vectors with their length proportional to the dog’s speed. Prepare: Since the dog accelerates for 30 m, the position dots will be successively farther apart and the velocity vectors will increase for this part of the race. After the dog reaches its top speed (at 30 m), the position dots are uniformly spaced and the velocity vectors are all the same length. Solve:

Assess: While the dog is accelerating, the dot spacing and the velocity vector must increase, and they do. While the dog is traveling at a constant speed the dot spacing and the velocity vector must remain the same, and they do.

P1.44. Strategize: We interpret the described motion in a motion diagram. We assume equal time intervals between dots, and we draw velocity vectors with their length proportional to the watermelon’s speed. Prepare: The watermelon, represented as a particle, falls freely and speeds up during its downward motion along the y-direction. The velocity vectors are thus of increasing length. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Representing Motion

1-23

Solve:

Assess: As a result of the acceleration due to gravity, as the watermelon falls, its velocity and the distance between the position dots increases. This is the case in the figure shown.

P1.45. Strategize: We will draw the motion diagram with velocity vectors initially pointing down an incline and then leveling off to travel over flat ground. Prepare: The length of velocity arrows and spacing between dots will increase as the skater rolls down the incline. Once the skater reaches the end of the incline, the skater will move at a constant speed. So on level ground the arrows should be a constant length and the dots should be equally-spaced. Solve:

Assess: We expect the speed to increase as the skater moves down the incline, and then remain steady. This is reflected in the diagram above.

P1.46. Strategize: We interpret the described motion in a motion diagram. We assume equal time intervals between dots, and we draw velocity vectors with their length proportional to the skater’s speed. Prepare: The skater moves along the x-axis. She slows down or has decreasing velocity vectors during a patch of rough ice. She has constant velocity vectors before the rough patch begins and after the rough patch ends, that is, velocity vectors are of the same length.

1-24

Chapter 1

Solve:

Assess: Before and after the rough ice, the velocity vector is constant in length and the position dots are uniformly spaced. Since the skater is traveling slower after the rough ice than before, the velocity vectors after the rough ice are shorter than they are before the rough ice and the position dots are closer together after the rough ice than they are before the rough ice. During the rough ice section, the velocity vector decreases in length and the dot position gets closer together.

P1.47. Strategize: We interpret the described motion in a motion diagram. We assume equal time intervals between dots, and we draw velocity vectors with their length proportional to the eland’s speed. Prepare: Since the eland has a positive velocity but is slowing down, the velocity will decrease to zero and the spacing between the position dots will decrease. The velocity vector at each position on the way up has the same magnitude but opposite direction as the velocity at each position on the way down. Solve:

Assess: On the way up, the velocity vector decreases to zero as it should and the spacing between the position dots decreases as it should. The magnitude of the velocity vector at any position is the same on the way up as it is on the way down. This allows us to conclude that the figure is correct.

P1.48. Strategize: We interpret the described motion in a motion diagram. We assume equal time intervals between dots, and we draw velocity vectors with their length proportional to the ball’s speed. Prepare: Represent the ball as a particle which is moving along the ramp defined as the x-axis. As the ball rolls up the ramp, it slows down. The velocity vector along the ramp is smaller than the vector along the floor. Solve:

Assess: While the ball is traveling across the horizontal section, the velocity vector and the spacing of the position dots is constant. As the ball makes the transition from the horizontal section to the ramp, there is a change in velocity. As the ball travels up the ramp the velocity vectors decrease in length and the position dots to get closer together. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Representing Motion

1-25

P1.49. Strategize: We interpret the described motion in a motion diagram. We assume equal time intervals between dots, and we draw velocity vectors with their length proportional to the car’s speed. Prepare: The motorist, represented as a particle, is moving along the x-axis. He slows down during braking. During his reaction time his velocity doesn’t change. Solve:

Assess: During the reaction time the velocity vector is constant in length and the position dots are uniformly spaced. During the braking process, the velocity vector decreases in length and the position dots get closer together.

P1.50. Strategize: Note any changes in velocity vectors, and make a physically plausible explanation for the changes. Prepare: Keep in mind that the dots represent the position of an object at equal time intervals and the vectors represent the velocity of the object at these times. Solve: Rahul was coasting on interstate highway I-44 from Tulsa to Springfield at 70 mph. Seeing an accident at a distance of 200 feet in front of him, he began to brake. What steady deceleration will bring him to a stop at the accident site? Assess: Since the position dots are initially equally spaced and the first few velocity vectors have the same length, this is consistent with Rahul initially traveling at a constant velocity. The fact that the dots get closer together and the velocity vectors get shorter is consistent with Rahul’s braking. The fact that there is no velocity vector associated with the last dot is consistent with the fact that he braked to a stop. P1.51. Strategize: Note any changes in velocity vectors, and make a physically plausible explanation for the changes. Prepare: Keep in mind that the dots represent the position of an object at equal time intervals and the vectors represent the velocity of the object at these times. Solve: Reema passes 3rd Street doing 40 mph, slows steadily to the stop sign at 4th Street, stops for 1 s, then speeds up and reaches her original speed as she passes 5th Street. If the blocks are 50 m long, how long does it take Reema to drive from 3rd Street to 5th Street? Assess: The statement that Reema slows to a stop in one block and regains her initial velocity in one block is consistent with the symmetry of the position dots and the velocity vectors about the stop position. P1.52. Strategize: Note any changes in velocity vectors, and make a physically plausible explanation for the changes. Prepare: Keep in mind that the dots represent the position of an object at equal time intervals and the vectors represent the velocity of the object at these times. Solve: A race car slows down from an initial speed of 100 mph to 50 mph in order to negotiate a tight turn. After making the 90° turn the car accelerates back up to 100 mph in the same time it took to slow down. Assess: The statement that the car slows down and speeds up in the same time is consistent with the symmetry of the spacing of the position dots and the length of the velocity vectors of the car going into and coming out of the curve. The statement that the car takes the curve at a constant speed is consistent with the fact that all the velocity vectors have the same length as the car negotiates the curve. P1.53. Strategize: Note any changes in velocity vectors, and make a physically plausible explanation for the changes. Prepare: Keep in mind that the dots represent the position of an object at equal time intervals and the vectors represent the velocity of the object at these times. We can construct a situation to match the motion diagram. Solve: A bowling ball is at rest at the top of an incline. You nudge the ball giving it an initial velocity and causing it to roll down an incline. At the bottom of the incline it bounces off a sponge and travels back up the incline until it stops. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 1

Assess: The statement that you give the ball an initial velocity is consistent with the fact that the start position dot has a velocity vector. The statement that the ball rolls down the incline is consistent with the fact that the dots are getting farther apart and the velocity vectors are increasing in length. The statement that the ball bounces off a sponge is consistent with the fact that ball does not bounce back to its original position.

P1.54. Strategize: Note any changes in velocity vectors, and make a physically plausible explanation for the changes. Prepare: Keep in mind that the dots represent the position of an object at equal time intervals and the vectors represent the velocity of the object at these times. We can construct a situation to match the motion diagram. Solve: Phil is standing on a third floor balcony, 10 m above the ground. He throws a ball with a speed of 20 m/s at an upward-angle of 30. Where does the ball hit the ground? Assess: The statement that the ball is kicked from the third floor and lands on the ground is consistent with the difference in heights of the start and end positions. Knowing that the ball slows down on the way up and speeds up on the way down is consistent with the fact that the position dots get closer together and the velocity vectors get shorter on the way up and the position dots get farther apart and the velocity vectors longer on the way down. P1.55. Strategize: This is an estimation problem, so many different answers are possible. Prepare: We must first estimate a human lifespan, and then do a series of unit conversions. A human lifespan depends heavily on where that human lives and on risk factors. For humans with ready access to medicines and without serious risk factors, at lifespan of 80 years is reasonable. We use this as our estimate. Solve: Converting 80 years into units of seconds, we find  365 days  24 hrs   3600 s  9 80 yr     = 2.5 10 s  1 yr  1 day   1 hr  Assess: We expect an extremely large number, since there are already thousands of seconds in just a single hour. This answer is reasonable.

P1.56. Strategize: We are asked for the average speed, which is given by speed =

distance . It is very important to time

1 ( v1 + v2 ) , because there is no reason to think that the two speeds will be equally important. 2 Prepare: We can determine the amount of time required for each leg of the journey, and then combine our information to determine the total distance and total time. Solve: For the first leg of the journey: v1 = d1 / t1  t1 = d1 / v1 = ( 25 mi ) / ( 55 mi/hr ) = 0.455 hr, and for the

realize that vav 

second leg of the journey, we have v2 = d2 / t2  t2 = d2 / v2 = (15 mi ) / ( 70 mi/hr ) = 0.214 hr Now the average speed for the entire trip is vav = d total / ttotal =

( 25 mi ) + (15 mi ) = 60 mi/hr. ( 0.455 hr ) + ( 0.214 hr )

Assess: We do expect the average speed to be somewhere between the two speeds at which Joseph traveled. It makes sense that this is the case. d to determine the new speed required. t Prepare: We know that the distance between Evan and his grandmother’s house does not change. But his speed and d the duration of the trip will both be different today than on a normal day. Thus we will write out vav,norm = tnorm

P1.57. Strategize: We will use the relationship vav =

and vav,today =

d ttoday

separately and then relate the two.

Solve: From the information about ordinary days, we know

 1 hr  d = vav,normtnorm = ( 55 mi/hr )( 25 min )   = 22.9 mi  60 min 

Representing Motion

1-27

Now we can use this distance, and the fact that Evan must make the trip in 5 minutes less than usual to write vav,today =

d ttoday

( 22.9 mi )  60 min  = 69 mph. ( 20 min )  1 hr 

Assess: Since Evan has to make the trip in 25% less time than usual, it makes perfect sense that his average speed must be 25% greater than usual.

P1.58. Strategize: We want to determine one average speed that Gretchen can run the entire time that is equivalent d total . ttotal Prepare: We can consider the two segments of the race during which Gretchen ran at a constant speed, separately. d d That way, we can write vav,1 = 1 and vav,2 = 2 , and then combine information to determine t2 t1 d d + d2 vav,total = total = 1 . ttotal t1 + t2 Solve: We write the expression for the average speed for the whole trip and then write the two relevant times in terms of the given speeds and distances.

to vav =

vav,total =

d total d +d d1 + d 2 = 1 2 = ttotal t1 + t2 ( d1 / vav,1 ) + ( d 2 / vav,2 )

4.0 103 m ) + (1.0 103 m ) ( vav,total = = 4.8 m/s (( 4.0 103 m ) / (5.0 m/s )) + ((1.0 103 m ) / ( 4.0 m/s ))

Assess: It makes sense that the average speed for the entire trip would be between the two speeds Gretchen ran during the race.

P1.59. Strategize: We will use the expression for average speed vav =

d and apply it to each leg of the swim t

(against the current, and with the current). Prepare: In order to calculate anything about the time, we must consider how the required time will be affected d d when we swim with the current: twith = . The distance is the , and against the current: tagainst = v + vw v − vw same either way. But we also know that we want to compare the total time we get from this to the time required in 2d . Here the factor of two came from the fact that we don’t need to consider two separate legs of still water: tstill = v the swim in this case. We can just write down the total two-way time. Solve: We want to compare twith + tagainst to tstill , so let us calculate the ratio

twith + tagainst tstill

d d +  v + vw v − vw 1  1 1 = =  +  2d / v 2  1 + vw / v 1 − vw / v 

 1 1 1 =  +  = 1.093 2  1 + ( 0.52 m/s ) / (1.78 m/s ) 1 − ( 0.52 m/s ) / (1.78 m/s )  This means the trip takes 9.3% longer taking the current into account, than it would in still water. Assess: This is a reasonable fractional difference, given that the speed of the water is significantly smaller than the swimming speed in still water.

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Chapter 1

P1.60. Strategize: This is a simple unit conversion. Prepare: The speed of the glacier has been given in feet per year. It is just a matter of unit conversion to get it into m/s. Solve: The speed of the glacier in m/s is determined as follows:  105 ft   12 in  254 cm   1 m   1 yr  v= = 10  10−6 m/s     7   yr   ft  1 in   100 cm   316  10 s 

Assess: All of the unit conversions are correct, after units are canceled we obtain the desired units (m/s) and we are expecting a very small number.

P1.61. Strategize: Prepare: Knowing that speed is distance divided by time, the distance is the circumference of a circle of radius 93,000,000 miles and the time is one year, we can determine the speed of the earth orbiting the sun. In order to get an answer in m/s, some unit conversion will be required. Solve: The speed of the earth in its orbit about the sun may be determined by v=

  161  103 m  distance 2 r 2 (93  107 mi)  1 yr 4 = =  = 30  10 m/s   7 time t 1 yr 1 mi 3  16  10 s   

Assess: All of the unit conversions are correct, after units are canceled we obtain the desired units (m/s) and we are expecting a large number.

P1.62. Strategize: This problem is mostly a unit conversion. We will also make use of the given growth rate of the shark. Prepare: The length of the shark can be converted to centimeters, and then the length can be related to an age using the given growth rate. The increase in length during the shark’s lifetime can be 14 ft. Solve: Starting with the increase in length, we have

(14 ft ) 

12 in  2.54 cm  1 yr     = 430 yrs.  1 ft  1 in  1 cm 

Assess: Although this is an incredible age, it is consistent with the statement that the Greenland shark is thought to be the longest-living vertebrate on Earth. d over any period when the average speed is given. t Prepare: We can apply the above expression to the uphill segment, the downhill segment.

P1.63. Strategize: We will use the equation vav =

vav,uphill =

d uphill tuphill

vav,downhill =

ddownhill , and we also note ttotal = tdownhill + tuphill tdownhill

We know both distances, the total time and the average speed for the uphill leg. We can relate the speeds, distances and times for each segment and solve for the unknown average speed on the downhill leg. d uphill Solve: Starting with ttotal = tdownhill + tuphill , and inserting tuphill = we can obtain vav,uphill tdownhill = ttotal −

d uphill vav,uphill

= ( 2,845 s ) −

( 4.6 mi )

 3600 s 

(8.75 mi/hr )  1 hr 

= 952 s

Now that we know the time, we simply use

vav,downhill =

ddownhill 6.9 mi  3600 s  =   = 26 mph tdownhill 952 s  1 hr 

Assess: This is much faster than the uphill leg of the trip, which is to be expected.

Representing Motion

1-29

P1.64. Strategize: We simply calculate the speed using Equation 1.1, and compare it to the purported value. Prepare: According to her speedometer, Shannon is traveling 70 mph. Knowing that mile markers are 1 mi apart and knowing that she travels the distance between two markers in 54 s, we can determine her speed and compare it to the value from the speedometer to see if her speedometer is accurate. Solve: Her actual speed in mph may be determined by v=

distance  1 mi   36  102 s  =  = 67 mph  time hr  54 s   

Since Shannon’s speedometer reads 70 mph and her actual speed is 67 mph, we conclude that her speedometer is not accurate. It shows a value that is too high. Assess: This is a reasonable answer—your speedometer is intended to give you a good estimate to your actual speed. Her speedometer is reading within 5% of the true value and rumor has it that the highway patrol will allow up to 10% on a good day. A well-calibrated speedometer could be this much off by changing the size of the tires.

P1.65. Strategize: Distance is the quantity measured by the odometer and depends on the path taken. The displacement is the difference between two locations, and is independent of the path taken between the two endpoints. The speed is the distance divided by the time interval. Prepare: Knowing that speed is distance divided by time, the distance is the speed multiplied by the time. 15 min = ¼ h. Solve: (a) The distance traveled during the ¼ hour is distance = speed  time = (100 km/h)(025 h) = 25 km (b) Since the circumference of the track is 12.5 km, then the car goes completely around the track exactly twice in covering 25 km. Hence, the displacement from the initial position is 0 km. (c) The speed of the car is

km  1000 m  1 h  1 min   v = 100     = 28 m/s h  1 km  60 min  60 s   Assess: 28 m/s seems like a reasonable speed for a fast car.

P1.66. Strategize: This problem involves estimation, such that many answers are possible. Prepare: Knowing the speed the signal travels (approximately 25 m/s) and estimating the distance from your brain to your hand to be about 1.0 m, we can determine the transmission time. Some unit conversion will be required to get the answer in ms. Solve: The transmission time for the signal may be determined by t=

distance  1 m   103 ms  =  = 40 ms  speed  25 m/s   s 

Assess: When you touch something very hot, it takes a fraction of a second to remove your hand. Keeping in mind, that in this case the signal must make a round trip, the answer, while very small, seems reasonable.

P1.67. Strategize: This is a straightforward application of the definition of speed. Prepare: Knowing the distance and time to travel that distance, we can determine the speed. Since we lack detailed information about the flight, and knowing that the flight is made by a bird, it is conceivable that its actual path darted back and forth and maybe even backtracked at times. If that is the case its actual distance of travel and hence average speed would be larger. Solve: (a) The minimum average speed of the albatross may be determined by v=

distance  12  103 km   1 day  0621 mi  =  = 22 mph  14 day   24 hr  time  1 km   

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Chapter 1

(b) The average speed of the bird is 22 mph if it flies in a straight line between the end points. If the bird deviates from this line, the average speed will have to be greater than 22 mph in order to have the 1200 km displacement in 1.4 days. Assess: You can ride a bike 10 to 15 mph and while you are riding your bike, birds easily fly past you. In light of that this is a reasonable answer.

P1.68. Strategize: This question involves unit conversion and the definition of speed. Prepare: We will need to convert to SI units throughout the problem. Solve: (a) The length is given as 2  m = 2  10−6 m (using the data from Table 1.2) (b) The diameter is 1  m = 110−6 m (using the data from Table 1.2) (c) The mass is given as 1 10−12 g = (1  10−12 g)(1 kg/1000 g) = 1 10−15 kg (d) The length of the bacterium’s DNA is given as 700 times longer than the bacterium’s length. Therefore the length of DNA is (700)(2  10−6 m) = 1  10−3 m We need to convert this to millimeters.  1000 mm  1  10−3 m = 1  10−3 m   = 1 mm  1m 

(e) The organism travels at 20 µm/s. Assume this is given to two significant figures. Converting to m/s, v = 20 µm/s = 2.0  10−5 m/s.

A day is

 24 h  60 min  60 s  1 day = (1 day)     = 86,400 s  d  h  min  So the bacterium travels

m  x = vt =  20  10−5  (86 400 s) = 17 m s  Assess: Use the method of multiplying by one to help keep track of multiple unit conversions.

P1.69. Strategize: This problem involves displacement, which is the difference between initial and final positions. We also distinguish between speed and velocity. Since velocity has a direction associated with it, in order for two segments to describe the same velocity, they would need to be the same length and the same direction. For them to describe the same speed, they only need to have the same length (since time intervals are fixed). Prepare: Assume that the bacterium moves along the path to consecutive letters. Solve: (a) The displacements in segments AB and CD are the same (five right and one up). No other pairs appear to be the same. (b) The problem explicitly stated that the bacteria move at a constant speed, so the answer is all of the segments. (c) Since the displacements in segments AB and CD are the same (and the bacterium had the same speed in both segments, i.e., t is the same for both segments), then the velocity is the same in those two segments. Since no other pairs of segments have the same direction, they can’t have the same velocity. Assess: Remember that both displacement and velocity are vectors, so the direction matters. Only the length (magnitude) matters with speed since it is a scalar. P1.70. Strategize: We can relate vertical and horizontal distances to the given angle using trigonometry. Prepare: The best way to prepare for this problem is to draw a diagram. In the triangle, the height h of the tree that we want to know is the side opposite to the given angle.

Representing Motion

1-31

Solve:

opposite adjacent opposite = adjacent  tan  tan  =

h = 52 m  tan30 = 30 m Assess: The shadow is longer than the height of the tree because the sun is low in the sky.

P1.71. Strategize: We can relate the distances given to the angle using trigonometry. Prepare: Draw a diagram of the situation. Note the right triangle.

Solve: Use knowledge about right triangles from trigonometry. (a) The angle of the path below the horizontal is

 360 m   = 21  920 m 

 = arctan  (b) The distance d covered is

d = (360 m)2 + (920 m)2 = 988 m which should be reported as 990 m to two significant figures. We keep one more significant figure to use in the next step. (c) The seal’s speed is

distance  988 m  1 min  =   = 41 m/s time  40 min  60 s 

Assess: Because the seal descended less than the horizontal distance we expect the angle to be less than 45 degrees.

P1.72. Strategize: This problem involves a displacement with components in two orthogonal directions. We will call east the + x axis and north the + y direction. Prepare: Erica’s displacement 25 north of east (which we will call d1 ) has both an x component and a y component, but the second leg (which we will call d2 ) has only a y component. We must determine the displacement in each orthogonal direction, and then use the Pythagorean Theorem to combine them and find the straight-line distance. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 1

Solve: For the total displacement eastward, we have x = d1 cos ( ) = ( 5.2 mi ) cos ( 25) = 4.71 mi

And for the displacement northward, we have y = d1 sin ( ) + d2 = ( 5.2 mi ) sin ( 25) + ( 4.0 mi ) = 6.20 mi

The total straight-line distance is given by

( x )2 + ( y )2 = ( 4.71mi )2 + ( 6.20 mi )2 = 7.8 mi

Assess: This is a reasonable distance. Note that this is less than the sum of the distances of the two legs, which it should be.

P1.73. Strategize: We can relate the given distance and angle to the required distance using trigonometry. Prepare: Draw a diagram of the situation. Note the right triangle.

Solve: Use knowledge about right triangles from trigonometry. (a) The horizontal distance x is x=

50 m = 2166 m tan13

This should be reported as 220 m to two significant figures. (b) The distance d covered is

d = (50 m)2 + (2166 m)2 = 2223 m which should be reported as 220 m to two significant figures. We keep two more significant figures to use in the next step. (c) The time it takes the shark is

time =

distance  2223 m  =  = 260 s speed  085 m/s 

Assess: 260 s is just over 4 min, which is impressive but reasonable.

P1.74. Strategize: Because the directions of motion are orthogonal, we can use the Pythagorean Theorem to add the two displacements. Prepare: Draw a right triangle with legs of length 3 and 4. The hypotenuse will have a length of 5. While we are given the time of flight for the legs, we are not actually told what the speed is, so we don’t really know the exact distance, but since the speed is constant we know the ratios of the sides of the triangle are as given.

Solve: The bird will fly for 3.0 min + 4.0 min + 5.0 min = 12.0 min. Assess: The diagram explains everything. The third side of the triangle is less than the sum of the first two sides.

Representing Motion

1-33

P1.75. Strategize: We can relate the speed to the displacement, easily. The two legs of the displacement are in orthogonal directions, such that we can use the Pythagorean Theorem. Prepare: Since during part of the motion John is traveling north and then turns east, the rules of vector addition will be used to determine the net displacement. Solve: (a) The vectors form a right triangle. See the following vector diagram.

The length of the net displacement vector is 2

AC = AB + BC = (100 km)2 + (100 km)2 = 141 km (b) Jane walks along John’s net displacement vector, so she only travels 1.41 km, while John travels a total distance of 2.00 km. Since he travels at 1.50 m/s during the entire stroll, the time John takes to get to his destination is

tJohn =

2000 m = 133  103 s 150 m/s

For Jane to walk 1.41 km in this time, her velocity would need to be

vJane =

1410 m 133 103 s

= 106 m/s

Assess: Jane must walk slower than John to walk the shorter distance in the same time, so the answer makes sense. For displacements in different directions you must use the law of vector addition.

P1.76. Strategize: This is a simple calculation of growth speed, given the change in height and the time interval. Prepare: Assume that the growth is steady, at least for the first three years. We will subtract any two heights and divide by the corresponding time interval. Solve: v=

30 ft − 12 ft 18 ft ft = =9 3 yr − 1 yr 2 yr yr

So the correct answer is B. Assess: You should get the same answer by choosing other values, such as year 2 and year 1. Try it.

P1.77. Strategize: This is a simple unit conversion problem. Prepare: We must convert ft/yr to m/s. Solve: The tree grows at 9 ft/yr, so 9

ft  9 ft   0305 m  1 yr  1 day  1 h  1 min  −8 =      = 9  10 m/s  yr  yr  1 ft  365 days  24 h  60 min  60 s 

The correct answer is A. Assess: Use the method of multiplying by one to keep track of conversion factors. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 1

P1.78. Strategize: Since we are given the vertical height of the tree and the horizontal distance from the tree, we have a right triangle and can use trigonometry. Prepare: Draw a diagram and label the sides and the angle you want to know.

Solve:

tan  =

opposite adjacent  opposite  −1  30 ft   = 63  = tan   adjacent   15 ft 

 = tan −1 

The correct answer is A. Assess: Imagine holding your protractor to your diagram and convince yourself that 63 is in the right ballpark.

MOTION IN ONE DIMENSION

Q2.1. Reason: The elevator must speed up from rest to cruising velocity. In the middle will be a period of constant velocity, and at the end a period of slowing to a rest. The graph must match this description. The value of the velocity is zero at the beginning, then it increases, then, during the time interval when the velocity is constant, the graph will be a horizontal line. Near the end the graph will decrease and end at zero.

Assess: After drawing velocity-versus-time graphs (as well as others), stop and think if it matches the physical situation, especially by checking end points, maximum values, places where the slope is zero, etc. This one passes those tests.

Q2.2. Reason: (a) The sign conventions for velocity are in Figure 2.7. The sign conventions for acceleration are in Figure 2.27. Positive velocity in vertical motion means an object is moving upward. Negative acceleration means the acceleration of the object is downward. Therefore the upward velocity of the object is decreasing. An example would be a ball thrown upward, before it starts to fall back down. Since it’s moving upward, its velocity is positive. Since gravity is acting on it and the acceleration due to gravity is always downward, its acceleration is negative. (b) To have a negative vertical velocity means that an object is moving downward. The acceleration due to gravity is always downward, so it is always negative. An example of a motion where both velocity and acceleration are negative would be a ball dropped from a height during its downward motion. Since the acceleration is in the same direction as the velocity, the velocity is increasing.

Assess: For vertical displacement, the convention is that upward is positive and downward is negative for both velocity and acceleration. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2-1

2-2

Chapter 2

Q2.3. Reason: Where the rings are far apart the tree is growing rapidly. It appears that the rings are quite far apart near the center (the origin of the graph), then get closer together, then farther apart again.

Q2.4. Reason: Call “up” the positive direction. Also assume that there is no air resistance. This assumption is probably not true (unless the rock is thrown on the moon), but air resistance is a complication that will be addressed later, and for small, heavy items like rocks no air resistance is a pretty good assumption if the rock isn’t going too fast. To be able to draw this graph without help demonstrates a good level of understanding of these concepts. The velocity graph will not go up and down as the rock does—that would be a graph of the position. Think carefully about the velocity of the rock at various points during the flight. At the instant the rock leaves the hand it has a large positive (up) velocity, so the value on the graph at t = 0 needs to be a large positive number. The velocity decreases as the rock rises, but the velocity arrow would still point up. So the graph is still above the t axis, but decreasing. At the tippy-top the velocity is zero; that corresponds to a point on the graph where it crosses the t axis. Then as the rock descends with increasing velocity (in the negative, or down, direction), the graph continues below the t axis. It may not have been totally obvious before, but this graph will be a straight line with a negative slope.

Assess: Make sure that the graph touches or crosses the t axis whenever the velocity is zero. In this case, that is only when it reaches the top of its trajectory and the velocity vector is changing direction from up to down. It is also worth noting that this graph would be more complicated if we were to include the time at the beginning when the rock is being accelerated by the hand. Think about what that would entail.

Motion in One Dimension

2-3

Q2.5. Reason: Let t0 = 0 be when you pass the origin. The other car will pass the origin at a later time t1 and passes you at time t 2 .

Assess: The slope of the position graph is the velocity, and the slope for the faster car is steeper.

Q2.6. Reason: The plot shows the x − component of the velocity as a function of time, not the position as a

function of time. When the lines A and B cross, this indicates that two objects have the same x − component of their velocity at that moment, not that they are in the same position. Of course, since we do not know where the objects started, it would be possible for the two objects to also be passing each other at that instant, there is simply no reason to think that based on the plot. So, Zach’s statement might technically be correct, but it does not capture any of the relevant features of the plot. So we reject his example. Let us ignore the fact that people often refer to the vertical direction as y, and assume that Victoria has chosen to call the vertical direction x (which is fine, just unconventional). The slope of either line indicates the change in velocity over time, in the x direction, which is called the acceleration in the x direction ax. When a rock is thrown, once it leaves a person’s hand gravity is always accelerating it downward. The vertical component of velocity may be upward or downward, but the acceleration is always downward. Line A has a positive acceleration and line B has a negative acceleration. It is not possible for these to both describe rocks in flight. We also reject Victoria’s example. Assess: An example of a scenario to accompany the plot would be: Car A was rolling backwards, but is now rolling forward and speeding up. Car B was moving forward, but slowed down and is now going in reverse. As Car A sped up and Car B slowed down, at one point they had the same speed.

Q2.7. Reason: A predator capable of running at a great speed while not being capable of large accelerations could overtake slower prey that were capable of large accelerations, given enough time. However, it may not be as effective as surprising and grabbing prey that are capable of higher acceleration. For example, prey could escape if the safety of a burrow were nearby. If a predator were capable of larger accelerations than its prey, while being slower in speed than the prey, it would have a greater chance of surprising and grabbing prey, quickly, though prey might outrun it if given enough warning. Assess: Consider the horse-man race discussed in the text. Q2.8. Reason: We will neglect air resistance, and thus assume that the ball is in free fall. (a) − g After leaving your hand the ball is traveling up but slowing, therefore the acceleration is down (i.e., negative). (b) -g At the very top the velocity is zero, but it had previously been directed up and will consequently be directed down, so it is changing direction (i.e., accelerating) down. (c) -g Just before hitting the ground it is going down (velocity is down) and getting faster; this also constitutes an acceleration down. Assess: As simple as this question is, it is sure to illuminate a student’s understanding of the difference between velocity and acceleration. Students would be wise to dwell on this question until it makes complete sense. Q2.9. Reason: Consider the ball thrown upward. The path from Janelle’s hand to its peak is symmetric to the path back from the peak to Janelle’s hand. That means that whatever the initial upward speed was, when the ball returns and passes Janelle on its way down, it will have that same speed, just directed downward now. From that moment on, the trip down to Michael is exactly the same as for the ball thrown downward. Thus the two balls will be moving at the same speed when they reach Michael.

2-4

Chapter 2

Assess: The ball initially thrown downward will certainly reach Michael first. But we are not asked about the time required or the average velocity. We are only asked about the speed at the moment the balls reach Michael. So this makes sense.

Q2.10. Reason: (a) Sirius the dog starts at about 1 m west of a fire hydrant (the hydrant is the x = 0 m position) and walks toward the east at a constant speed, passing the hydrant at t = 1.5 s. At t = 4 s Sirius encounters his faithful friend Fido 2 m east of the hydrant and stops for a 6-second barking hello-and-smell. Remembering some important business, Sirius breaks off the conversation at t = 10 s and sprints back to the hydrant, where he stays for 4 s and then leisurely pads back to his starting point. (b) Sirius is at rest during segments B (while chatting with Fido) and D (while at the hydrant). Notice that the graph is a horizontal line while Sirius is at rest. (c) Sirius is moving to the right whenever x is increasing. That is only during segment A. Don’t confuse something going right on the graph (such as segments C and E) with the object physically moving to the right (as in segment A). Just because t is increasing doesn’t mean x is. (d) The speed is the magnitude of the slope of the graph. Both segments C and E have negative slope, but C’s slope is steeper, so Sirius has a greater speed during segment C than during segment E. Assess: We stated our assumption (that the origin is at the hydrant) explicitly. During segments B and D time continues to increase but the position remains constant; this corresponds to zero velocity. Q2.11. Reason: There are five different segments of the motion, since the lines on the position-versus-time graph have different slopes between five different time periods. (a) A fencer is initially still. To avoid his opponent’s lunge, the fencer jumps backwards very quickly. He remains still for a few seconds. The fencer then begins to advance slowly on his opponent. (b) Referring to the velocities obtained in part (a), the velocity-versus-time graph would look like the following diagram.

Assess: Velocity is given by the slope of lines on position-versus-time graphs. See Conceptual Example 2.1 and the discussion that follows.

Q2.12. Reason: (a) A’s speed is greater at t = 1 s. The slope of the tangent to B’s curve at t = 1 s is smaller than the slope of A’s line. (b) A and B have the same speed just before t = 3 s. At that time, the slope of the tangent to the curve representing B’s motion is equal to the slope of the line representing A’s motion. Assess: The fact that B’s curve is always above A’s doesn’t really matter. The respective slopes matter, not how high on the graph the curves are.

Q2.13. Reason: (a) D. The steepness of the tangent line is greatest at D. (b) C, D, E. Motion to the left is indicated by a decreasing segment on the graph. (c) C. The speed corresponds to the steepness of the tangent line, so the question can be re-cast as “Where is the tangent line getting steeper (either positive or negative slope, but getting steeper)?” The slope at B is zero and is greatest at D, so it must be getting steeper at C. (d) A, E. The speed corresponds to the steepness of the tangent line, so the question can be re-cast as “Where is the tangent line getting less steep (either positive or negative slope, but getting less steep)?” (e) B. Before B the object is moving right and after B it is moving left. Assess: It is amazing that we can get so much information about the velocity (and even about the acceleration) from a position-versus-time graph. Think about this carefully. Notice also that the object is at rest (to the left of the origin) at point F. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Motion in One Dimension

2-5

Q2.14. Reason: (a) For the velocity to be constant, the velocity-versus-time graph must have zero slope. Looking at the graph, there are three time intervals where the graph has zero slope: segment A, segment D and segment F. (b) For an object to be speeding up, the magnitude of the velocity of the object must be increasing. When the slope of the lines on the graph is nonzero, the object is accelerating and therefore changing speed. Consider segment B. The velocity is positive while the slope of the line is negative. Since the velocity and acceleration are in opposite directions, the object is slowing down. At the start of segment B, we can see the velocity is +2 m/s, while at the end of segment B the velocity is 0 m/s. During segment E the slope of the line is positive which indicates positive acceleration, but the velocity is negative. Since the acceleration and velocity are in opposite directions, the object is slowing here also. Looking at the graph at the beginning of segment E the velocity is –2 m/s, which has a magnitude of 2 m/s. At the end of segment E the velocity is 0 m/s, so the object has slowed down. Consider segment C. Here the slope of the line is negative and the velocity is negative. The velocity and acceleration are in the same direction so the object is speeding up. The object is gaining velocity in the negative direction. At the beginning of that segment the velocity is 0 m/s, and at the end the velocity is –2 m/s, which has a magnitude of 2 m/s. (c) In the analysis for part (b), we found that the object is slowing down during segments B and E. (d) An object standing still has zero velocity. The only time this is true on the graph is during segment F, where the line has zero slope, and is along v = 0 m/s. The velocity is also zero for an instant at time t = 5 s between segments B and C. (e) For an object to be moving to the right, the convention is that the velocity is positive. In terms of the graph, positive values of velocity are above the time axis. The velocity is positive for segments A and B. The velocity must also be greater than zero. Segment F represents a velocity of 0 m/s. Assess: The slope of the velocity graph is the acceleration graph. Q2.15. Reason: This graph shows a curved position-versus-time line. Since the graph is curved the motion is not uniform. The instantaneous velocity, or the velocity at any given instant of time, is the slope of a line tangent to the graph at that point in time. Consider the graph below, where tangents have been drawn at each labeled time.

Comparing the slope of the tangents at each time in the figure above, the speed of the car is greatest at time C. Assess: Instantaneous velocity is given by the slope of a line tangent to a position-versus-time curve at a given instant of time. This is also demonstrated in Conceptual Example 2.4.

Q2.16. Reason: C. Negative, negative; since the slope of the tangent line is negative at both 1 and 2. Assess: The car’s position at 2 is at the origin, but it is traveling to the left and therefore has negative velocity in this coordinate system. Q2.17. Reason: The velocity of an object is given by the physical slope of the line on the position-versus-time graph. Since the graph has constant slope, the velocity is constant. We can calculate the slope by using Equation 2.1, choosing any two points on the line since the velocity is constant. In particular, at t1 = 0 s the position is x1 = 5 m. At time t2 = 3 s the position is x2 = 15 m. The points on the line can be read to two significant figures. The velocity is v=

Dx x2 - x1 15 m - 5 m 10 m = = = = +3.3 m/s Dt t2 - t1 3 s-0 s 3s

The correct choice is C. Assess: Since the slope is positive, the value of the position is increasing with time, as can be seen from the graph. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2-6

Chapter 2

Q2.18. Reason: We are asked to find the largest of four accelerations, so we compute all four from Equation 2.8: ax = A ax =

10 m/s = 2.0 m/s2 5.0 s

B ax =

5.0 m/s = 2.5 m/s2 2.0 s

C ax =

20 m/s = 2.9 m/s2 7.0 s

Dvx Dt

3.0 m/s = 3.0 m/s2 1.0 s The largest of these is the last, so the correct choice is D. Assess: A large final speed, such as in choices A and C, does not necessarily indicate a large acceleration. D ax =

Q2.19. Reason: The initial velocity is 20 m/s. Since the car comes to a stop, the final velocity is 0 m/s. We are given the acceleration of the car, and need to find the stopping distance. See the pictorial representation, which includes a list of values below.

An equation that relates acceleration, initial velocity, final velocity, and distance is Equation 2.13.

(vx )2f = (vx )2i + 2ax Dx Solving for Dx,

Dx =

(vx )2f - (vx )2i (0 m/s)2 - (20 m/s)2 = = 50 m 2ax 2(-4.0 m/s2 )

The correct choice is D. Assess: We are given initial and final velocities and acceleration. We are asked to find a displacement, so Equation 2.13 is an appropriate equation to use.

Q2.20. Reason: This is not a hard question once we remember that the displacement is the area under the velocityversus-time graph. The scales on all three graphs are the same, so simple visual inspection will attest that Betty traveled the furthest since there is more area under her graph. The correct choice is B. Assess: It is important to verify that the scales on the axes on all the graphs are the same before trusting such a simple visual inspection. In the same vein, it is important to realize that although all three cars end up at the same speed (40 m/s), they do not end up at the same place (assuming they started at the same position); this is nothing more than reiterating what was said in the Reason step above. On a related note, check the accelerations: Andy’s acceleration was small to begin with but growing toward the end, Betty’s was large at first and decreased toward the end, and Carl’s acceleration was constant over the 5.0 s. Mentally tie this all together.

Motion in One Dimension

2-7

Q2.21. Reason: The slope of the tangent to the velocity-versus-time graph gives the acceleration of each car. At time t = 0 s the slope of the tangent to Andy’s velocity-versus-time graph is very small. The slope of the tangent to the graph at the same time for Carl is larger. However, the slope of the tangent in Betty’s case is the largest of the three. So Betty had the greatest acceleration at t = 0 s. See the figure below.

The correct choice is B. Assess: Acceleration is given by the slope of the tangent to the curve in a velocity-versus-time graph at a given time.

Q2.22. Reason: Both balls are in free fall (neglecting air resistance) once they leave the hand, and so they will have the same acceleration. Therefore, the slopes of their velocity-versus-time graphs must be the same (i.e., the graphs must be parallel). That eliminates choices B and C. Ball 1 has positive velocity on the way up, while ball 2 never goes up or has positive velocity; therefore, choice A is correct. Assess: Examine the other choices. In choice B ball 1 is going up faster and faster while ball 2 is going down faster and faster. In choice C ball 1 is going up the whole time but speeding up during the first part and slowing down during the last part; ball 2 is going down faster and faster. In choice D ball 2 is released from rest (as in choice A), but ball 1 is thrown down so that its velocity at t = 0 is already some non-zero value down; thereafter both balls have the same acceleration and are in free fall. Q2.23. Reason: There are two ways to approach this problem, and both are educational. Using algebra, first calculate the acceleration of the larger plane. a=

 v 80 m/s = = 2.667 m/s 2 t 30 s

Then use that acceleration to figure how far the smaller plane goes before reading 40 m/s. (v x ) f 2 = ( v x ) i 2 + 2 a x  x   x =

(v x ) f 2 − ( v x ) i 2 (40 m/s) 2 = = 300 m 2a x 2(2.667 m/s 2 )

So choice A is correct. The second method is graphical. Make a velocity vs. time graph; the slope of the straight line is the same for both planes. We see that the smaller plane reaches 40 m/s in half the time that the larger plane took to reach 80 m/s. And we see that the area under the smaller triangle is ¼ the area under the larger triangle. Since the area under the velocity vs. time graph is the distance then the distance the small plane needs is ¼ the distance the large plane needs.

Assess: It seems reasonable that a smaller plane would need only ¼ the distance to take off as a large plane.

2-8

Chapter 2

Q2.24. Reason: The dots from time 0 to 9 seconds indicate a direction of motion to the right. The dots are getting closer and closer. This indicates that the object is moving to the right and slowing down. From 9 to 16 seconds, the object remains at the same position, so it has no velocity. From 16 to 23 seconds, the object is moving to the left. Its velocity is constant since the dots are separated by identical distances. The velocity-versus-time graph that matches this motion closest is B. Assess: The slope of the line in a velocity-versus-time graph gives an object’s acceleration. Q2.25. Reason: Let us call the vertically upward direction + y. We can determine the initial velocity by using Equation 2.13, with the minor adjustment of changing the subscripts from x to y. Then we have

( v ) = ( v ) + 2a y  ( v ) =  ( v ) − 2a y =  ( 2.8 m/s ) − 2 ( −9.8 m/s )(3.8 m ) = 9.1 m/s 2

Clearly, since the ball was thrown upward, we want the positive answer. So the correct answer is C. Assess: It makes sense that the initial upward component of velocity would have to be greater in magnitude than the final.

Q2.26. Reason: We can solve this in two steps. First, we can use Equation 2.12 to determine the acceleration of the car. Then we can use that acceleration to determine the time required to travel 30 m. The difference in time required to drive 30 m and to drive the initial 15 m will be the time required to drive the second 15 m. Let us call the direction of motion + x. The acceleration is given by  x = ( v x )i  t +

2 (15 m ) 1 2 x 2 ax ( t )  ax = = = 5.21 m/s 2 2 2 2 ( t ) ( 2.4 s )

Here, we have used the fact that the car started from rest. Now we consider the entire 30 m trip, and use the acceleration we just found:

1 2 x = ( vx )i t + ax ( t )  t = 2

2 x = ax

2 ( 30 m )

(5.21m/s ) 2

= 3.39 s

The entire 30 m takes 3.39 s, and the first 15 m took 2.4 s. This means the second 15 m segment takes 1.0 s. The correct answer is A. Assess: It makes sense that the second 15 m segment takes less time than the first 15 m segment, since the car already had gotten up to speed .

Q2.27. Reason: By definition the acceleration in the x direction is the rate of change of the velocity in the x direction. This means a x =  vx /  t =

( vx )f − ( vx )i (15 m/s ) − (5 m/s ) = = 2.5 m/s 2 . tf − ti ( 4 s) − (0 s)

The correct answer is B. Assess: Since the slope is positive, we expect a positive acceleration. The magnitude is also reasonable for the components of velocity given.

Q2.28. Reason: We can solve this with a straightforward application of Equation 2.13. One can easily obtain the acceleration from the slope, and this was the goal of Q2.27. The result is ax = 2.5 m/s 2 ; see the solution to Q2.27 for details.

( vx )f = ( vx )i + 2ax x  x = 2

( vx )f − ( vx )i 2

2a x

(15 m/s )f − (5 m/s )i 2

(

2 2.5 m/s 2

)

= 40 m

The correct answer is C. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Motion in One Dimension

2-9

Assess: The speed is never less than 5 m/s, so the distance traveled must be greater than ( 5 m/s )( 4 s ) = 20 m. The speed is never greater than 15 m/s, so the distance traveled must be less than (15 m/s )( 4 s ) = 60 m. Our answer lies directly between these limits, and is therefore very reasonable.

Problems P2.1. Strategize: The dots represent positions after fixed time intervals. Prepare: The car is traveling to the left toward the origin, so its position decreases with increase in time. Solve: (a) Time t (s) Position x (m) 0 1200 1 975 2 825 3 750 4 700 5 650 6 600 7 500 8 300 9 0 (b)

Assess: A car’s motion traveling down a street can be represented at least three ways: a motion diagram, positionversus-time data presented in a table (part (a)), and a position-versus-time graph (part (b)).

P2.2. Strategize: Review our sign conventions. Prepare: Position to the right of or above origin is positive, but to the left of or below origin is negative. Velocity is positive for motion to the right and for upward motion, but it is negative for motion to the left and for downward motion. Solve: Diagram

Position

Velocity

(a) (b) (c)

Negative Negative Positive

Positive Negative Negative

P2.3. Strategize: This is a position vs. time graph, so the x component of velocity is given by the slope. Prepare: The position graph has a shallow (negative) slope for the first 8 s, and then the slope increases. Solve: (a) The change in slope comes at 8 s, so that is how long the dog moved at the slower speed. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2-10

Chapter 2

(b)

Assess: We expect the sneaking up phase to be longer than the spring phase, so this looks like a realistic situation.

P2.4. Strategize: To get a position from a velocity graph we count the area under the curve. Prepare: We expect to draw a line on our position graph that has a positive slope whenever the x component of velocity is positive, and a negative slope when the x component of velocity is negative. Solve: (a)

(b) We need to count the area under the velocity graph (area below the x-axis is subtracted). There are 18 m of area above the axis and 4 m of area below. 18 m − 4 m =14 m. Assess: These numbers seem reasonable; a mail carrier could back up 4 m. It is also important that the problem state what the position is at t = 0, or we wouldn’t know how high to draw the position graph.

P2.5. Strategize: We want to produce a position vs. time plot in which the slope of x vs. t yields the vx vs. t plot we are given. Prepare: To get a position from a velocity graph we count the area under the curve. Solve: (a)

Motion in One Dimension

2-11

(b) We need to count the area under the velocity graph (area below the x-axis is subtracted). There are 12 m of area below the axis and 12 m of area above. 12 m − 12 m = 0 m. (c) A football player runs left at 3 m/s for 4 s, then cuts back to the right at 3 m/s for 2 s, then walks (continuing to the right) back to the starting position. Assess: We note an abrupt change of velocity from 3 m/s left to 3 m/s right at 4 s. It is also important that the problem state what the position is at t = 0, or we wouldn’t know how high to draw the position graph.

P2.6. Strategize: This problem can be broken down into steps during which the velocity is constant, such that (during any step) the relationship between displacement, velocity and time is simple. Prepare: Let us call east the positive x direction. During any period when the velocity is constant, we can use vx = x / t. We are told the time for the 5 minute stop, and for the 20 minute trip with the wind. We can use this to calculate the distance that Dylan covers, and then the time required for the return. Finally, we will add the times from each segment to obtain a total time. Solve: For the first part of the trip, when Dylan rides with the wind, we have:  15 mi  vx =  x /  t   x = v x  t =   ( 20 min ) = 5.0 mi  60 min 

Dylan must cover this same distance during the return trip, but this time he will be moving in the − x direction (meaning both the displacement in x and the x component of velocity will be negative). We have:  −10 mi  vx = x / t  t = x / vx = ( − 5.0 mi ) /   = 30 min  60 min 

The total time is then the sum of the 20 min, 5 min, and 30 min segments, or 55 minutes. Assess: It makes sense that the return trip takes 10 minutes longer than the first leg of the trip, since the wind is blowing against Dylan for the return trip.

P2.7. Strategize: We want to indicate position relative to the ground on the vertical axis of one plot, and time on the horizontal axis. Prepare: We assume the speed is roughly constant when the elevator is moving. We start with the position plot, and then we can determine the components of the velocity from the slope. Solve: The entire trip takes 24 s and 10 s are spent stopped. So the motion takes a total of 14 s. If we assume the elevator has the same speed when going upward as it does downward, then the total distance of 7 floors (5 up and then 2 down) corresponds to 2 s per floor. Thus, we expect it took 10 s to get up to the fifth floor, and then 4 s to go back down to the third floor. Clearly, the slope of this plot is either 2 m/s or 0 m/s. This allows us to complete the velocity vs. time plot as well.

Assess: Note that the sign of the velocity is accurately reflected in the slope of the position vs. time plot.

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Chapter 2

P2.8. Strategize: The slope of the position-versus-time graph at every point gives the velocity at that point. Prepare: Referring to Figure P2.8, the graph has a distinct slope and hence distinct velocity in the time intervals: from t = 0 to t = 20 s; from 20 s to 30 s; and from 30 s to 40 s. Solve: The slope at t = 10 s is v=

Dx 100 m - 50 m = = 2.5 m/s Dt 20 s

The slope at t = 25 s is

100 m - 100 m = 0 m/s 10 s

0 m - 100 m = -10 m/s 10 s

The slope at t = 35 s is

Assess: As expected a positive slope gives a positive velocity and a negative slope yields a negative velocity.

P2.9. Strategize: Ignoring air resistance, the horizontal component of the velocity should be constant. Prepare: Assume that the ball travels in a horizontal line at a constant v x . It doesn’t really, but if it is a line drive then it is a fair approximation. time =

distance 60 ft  1 mi  60 min  60 s  =   = 0.43 s speed 95 mih  5280 ft   1 h  1 min 

Assess: Just under a half second is reasonable for a major league pitch.

P2.10. Strategize: Ignoring air resistance, the horizontal component of the velocity should be constant. Prepare: Assume that the ball travels in a horizontal line at a constant v x . It doesn’t really, but if it is a line drive then it is a fair approximation. Solve: t =

x 43 ft  1 mi  3600 s  =    = 0.29 s vx 100 mi/h  5280 ft  1 h 

Assess: This is a short but reasonable time for a fastball to get from the mound to home plate.

P2.11. Strategize: We assume both drivers maintain a steady speed. Prepare: A visual overview of Alan’s and Beth’s motion that includes a pictorial representation, a motion diagram, and a list of values is shown below. Our strategy is to calculate and compare Alan’s and Beth’s time of travel from Los Angeles to San Francisco.

Motion in One Dimension

2-13

Solve: Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows:

x - xi Dx xf - xi = Þ tf = ti + f Dt tf - t i v

Using the known values identified in the pictorial representation, we find

(tf )Alan = (ti )Alan +

(xf )Alan - (xi )Alan 400 mile = 8:00 AM + = 8:00 AM + 8 hr = 4:00 PM v 50 miles/hour

(tf ) Beth = (ti ) Beth +

(xf ) Beth - (xi ) Beth 400 mile = 9:00 AM + = 9:00 AM + 6.67 hr = 3:40 PM 60 miles/hour v

(a) Beth arrives first. (b) Beth has to wait 20 minutes for Alan. Assess: Times of the order of 7 or 8 hours are reasonable in the present problem.

P2.12. Strategize: This problem involves constant-velocity motion. Prepare: Assume that Richard only speeds on the 125 mi stretch of the interstate. We then need to compute the times that correspond to two different speeds for that given distance. Rearrange Equation 1.1 to produce time =

distance speed

Solve: At the speed limit: time1 =

125 mi  60 min    = 115.4 min 65 mi/h  1 h 

time 2 =

125 mi  60 min    = 107.1 min 70 mi/h  1 h 

At the faster speed:

By subtracting we see that Richard saves 8.3 min. Assess: Breaking the law to save 8.3 min is normally not worth it; Richard’s parents can wait 8 min. Notice how the hours (as well as the miles) cancel in the equations.

P2.13. Strategize: Since each runner is running at a steady pace, they both are traveling with a constant speed. Prepare: Each runner must travel the same distance to finish the race. We assume they are traveling uniformly. We can calculate the time it takes each runner to finish using Equation 2.1. Solve: The first runner finishes in t1 =

x 5.00 km = = 0.417 h (vx )1 12.0 km/h

 60 min  Converting to minutes, this is (0.417 h)   = 25.0 min  1h  For the second runner

t 2 =

x 5.00 km = = 0.345 h (vx )2 14.5 km/h

Converting to seconds, this is  60 min  (0.345 h)   = 20.7 min  1h 

The time the second runner waits is 25.0 min – 20.7 min = 4.3 min Assess: For uniform motion, velocity is given by Equation 2.1.

2-14

Chapter 2

P2.14. Strategize: This problem involves constant-velocity motion. Prepare: We’ll do this problem in multiple steps. Rearrange Equation 1.1 to produce time =

distance speed

Use this to compute the time the faster runner takes to finish the race; then use distance = speed ´ time to see how far the slower runner has gone in that amount of time. Finally, subtract that distance from the 8.00 km length of the race to find out how far the slower runner is from the finish line.

Solve: The faster runner finishes in

8.00 km = 0.571 h 14.0 km/h

In that time the slower runner runs d = (11.0 km/h) ´ (0.571 h) = 6.29 km. This leaves the slower runner 8.00 km - 6.29 km = 1.71 km from the finish line as the faster runner crosses the line. Assess: The slower runner will not even be in sight of the faster runner when the faster runner crosses the line. We did not need to convert hours to seconds because the hours cancelled out of the last equation. Notice we kept 3 significant figures, as indicated by the original data.

P2.15. Strategize: This is a position vs. time plot, and we are asked for the top speed. So we are interested in the maximum slope in the given plot. Prepare: The slope of the path traced by the dots in the position vs. time plot is initially somewhat inclined, but then increases after the first two seconds. After that point near the two second mark (which we approximate as 1.9 s), the slope appears fairly constant. We can take this larger, sustained slope as the maximum speed. Solve: Since the speed after the dot at 1.9 s mark appears roughly constant, we can use the expression for the average speed over that interval: vx ,av =

x xf − xi (100 m ) − (10 m ) = = = 12 m/s t tf − ti ( 9.5 s ) − (1.9 s )

Assess: This is extremely fast. One can perform a simple check by noting that the average time for the entire run is (100 m ) / (9.5 s ) = 11 m/s. So the fact that we got a slightly higher speed for Usain Bolt’s maximum speed is reasonable.

Motion in One Dimension

2-15

P2.16. Strategize: Since she is running at a steady pace, this is a constant-velocity problem. Prepare: Assume v x is constant so the ratio

Dx is also constant. Dt

Solve: (a)

100 m 400 m  400 m  =  t = 18 s   = 72 s 18 s t  100 m  (b)

100 m 1.0 mi  1.0 mi  1609 m  =  t = 18 s    = 290 s = 4.8 min 18 s t  100 m  1.0 mi  Assess: This pace does give about the right answer for the time required to run a mile for a good marathoner.

P2.17. Strategize: In this problem the velocity is changing, but we can determine vx at a given time by looking at the slope of the x vs. t graph. Prepare: The graph in Figure P2.17 shows distinct slopes in the time intervals: 0 – 1 s, 1 s – 2 s, and 2 s − 4 s. We can thus obtain the velocity values from this graph using v = x/t. Solve: (a)

(b) There is only one turning point. At t = 2 s the velocity changes from +20 m/s to −10 m/s, thus reversing the direction of motion. At t = 1 s, there is an abrupt change in motion from rest to +20 m/s, but there is no reversal in motion. Assess: As shown above in (a), a positive slope must give a positive velocity and a negative slope must yield a negative velocity.

P2.18. Strategize: The distance traveled is the area under the v y graph. Prepare: Since the graph of v y vs. t is linear in each region, calculating the area under the line is simple. Solve: (a) The area of a triangle is

1 BH . 2  y = area =

1 1 BH = (0.20 s)(0.75 m/s) = 7.5 cm 2 2

(b) We estimate the distance from the heart to the brain to be about 30 cm.

Dt =

Dy 30 cm = = 4.0 beats v y 7.5 cm/beat

Assess: Four beats seems reasonable. There is some doubt that we are justified using two significant figures here. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2-16

Chapter 2

P2.19. Strategize: Displacement is given by the area under the a velocity vs. time graph. Prepare: In this case, the displacement is equal to the area under the velocity graph between ti and tf. We can find the car’s final position from its initial position and the area. Solve: (a) Using the equation xf = xi + area of the velocity graph between ti and tf, x2 s = 10 m + area of trapezoid between 0 s and 2 s 1 = 10 m + (12 m /s + 4 m/s)(2 s) = 26 m 2 x3 s = 10 m + area of triangle between 0 s and 3 s 1 = 10 m + (12 m /s)(3 s) = 28 m 2 x4 s = x3 s + area between 3 s and 4 s 1 = 28 m + (-4 m/s)(1 s) = 26 m 2 (b) The car reverses direction at t = 3 s, because its velocity becomes negative. Assess: The car starts at xi = 10 m at ti = 0. Its velocity decreases as time increases, is zero at t = 3 s, and then becomes negative. The slope of the velocity-versus-time graph is negative which means the car’s acceleration is negative and a constant. From the acceleration thus obtained and given velocities on the graph, we can also use kinematic equations to find the car’s position at various times.

P2.20. Strategize: This is an estimation problem, so a range of answers may be acceptable. vx is given by the slope of the x vs. graph. Prepare: To make the estimates from the graph we need to read the slopes from the graph. Lightly pencil in straight lines tangent to the graph at t = 2 s and t = 4 s. Then pick a pair of points on each line to compute the rise and the run. Solve: (a)

vx =

200 m = 67 m/s 4 s -1 s

vx =

400 m = 130 m/s 5 s-2 s

(b)

Assess: The speed is increasing, which is indeed what the graph tells us. These are reasonable numbers for a drag racer.

P2.21. Strategize: The graph in Figure P2.21 shows the horizontal component of velocity as a function of time. We know the acceleration is the rate of change of the velocity. So we can determine the acceleration using the slope of this graph. Prepare: We will use ax = vx/t. A linear decrease in velocity from t = 0 s to t = 2 s implies a constant negative acceleration. On the other hand, a constant velocity between t = 2 s and t = 4 s means zero acceleration. Solve:

Motion in One Dimension

2-17

P2.22. Strategize: Figure P2.22 shows the horizontal component of the velocity. We know the displacement in the horizontal direction is the area under this curve, and acceleration in the horizontal direction is the given by the slope of this graph. Prepare: To determine the displacement at a time t, we calculate the area between the velocity line and the time axis up to that time. To determine the horizontal component of the acceleration, we use a x = vx / t. Solve: (a)

(b) From the acceleration versus t graph above, ax at t = 3.0 s is +1 m/s2. Assess: Because the velocity was negative at first, the train was moving left. There is a turning point at 2 s.

P2.23. Strategize: Acceleration is the rate of change of velocity. We must draw a velocity vs. time graph in which the slope at a given point is equal to the value of the acceleration in the plot above. Prepare: We are told the initial speed of the object is 2.0 m/s. We can simply start drawing a line from that point with the appropriate slope, changing slopes at the appropriate times. Solve:

Assess: We can check our answer by calculating the velocity after a certain time and seeing if it matches the graph. Let us check the lowest point, which on our graph is –6.0 m/s and occurs at 4 s. Using Equation 2.11, we have (vx )f = (vx )i + axt = ( 2.0 m/s) + −2.0 m/s2 ( 4 s ) = −6.0 m/s, which is consistent.

(

)

P2.24. Strategize: Acceleration is given by the slope of the velocity vs. time graph. Prepare: We calculate the acceleration from the rise and the run for each straight line segment. Solve: Speeding up: ay =

Dv y Dt

0.75 m/s = 15 m/s2 0.05 s

2-18

Chapter 2

Slowing down:

ay =

Dv y Dt

-0.75 m/s = -5 m/s2 0.15 s

Assess: Indeed the slope looks three time steeper in the first segment than in the second. These are pretty large accelerations.

P2.25. Strategize: From a velocity-versus-time graph we find the acceleration by computing the slope. Prepare: We will compute the slope of each straight-line segment in the graph. ax =

(vx )f - (vx )i tf - ti

The trickiest part is reading the values off of the graph. Solve: (a)

ax =

5.5 m/s - 0.0 m/s = 6.1 m/s2 0.9 s - 0.0 s

ax =

9.3 m/s - 5.5 m/s = 2.5 m/s2 2.4 s - 0.9 s

(b)

(c)

ax =

10.9 m/s - 9.3 m/s = 1.5 m/s2 3.5 s - 2.4 s

Assess: This graph is difficult to read to more than one significant figure. I did my best to read a second significant figure but there is some estimation in the second significant figure. It takes Carl Lewis almost 10 s to run 100 m, so this graph covers only the first third of the race. Were the graph to continue, the slope would continue to decrease until the slope is zero as he reaches his (fastest) cruising speed. Also, if the graph were continued out to the end of the race, the area under the curve should total 100 m.

P2.26. Strategize: In reality, biological systems rarely move with constant acceleration. But we will assume constant acceleration over this very short time interval. Prepare: Use the definition of acceleration. Also, 60 ms = 0.060 s. Solve: ay =

Dv y Dt

3.7 m/s = 62 m/s2 0.060 s

Assess: Frogs are quite impressive! Humans can’t jump with this kind of acceleration.

P2.27. Strategize: We will assume constant accelerations for both animals. Prepare: We can calculate acceleration from Equation 2.8: Solve: For the gazelle:  v  13 m/s ( ax ) =  x  = = 4.3 m/s 2  t  3.0 s

For the lion:  v  9.5 m/s ( ax ) =  x  = = 9.5 m/s 2 1.0 s  t 

Motion in One Dimension

2-19

For the trout:  v  2.8 m/s ( ax ) =  x  = = 23 m/s 2  t  0.12 s

The trout is the animal with the largest acceleration. Assess: A lion would have an easier time snatching a gazelle than a trout.

P2.28. Strategize: We will assume constant acceleration, such that we can use kinematic equations. Prepare: Acceleration is the rate of change of velocity. ax =

Dvx Dt

Where Dvx = 4.0 m/s and Dt = 0.11 s. We will then use that acceleration to compute the final position after the strike:

xf =

1 a (Dt)2 2 x

where we are justified in using the special case because (vx )i = 0.0 m/s and xi = 0 m. Solve: (a)

ax =

Dvx 4.0 m/s = = 36 m/s2 Dt 0.11 s

(b)

xf =

1 1 ax (Dt)2 = (36 m/s2 )(0.11s)2 = 0.22 m 2 2

Assess: The answer is remarkable but reasonable. The pike strikes quickly and so is able to move 0.22 m in 0.11 s, even starting from rest. The seconds squared cancel in the last equation.

P2.29. Strategize: This problem consists of unit conversion, and application of the definition of acceleration. Note that since the acceleration is constant, we are also free to use kinematic equations. Prepare: First, we will convert units: 60

miles 1 hour 1609 m ´ ´ = 26.8 m/s hour 3600 s 1 mile

We also note that g = 9.8 m/s2. Because the car has constant acceleration, we can use kinematic equations. Solve: (a) For initial velocity vi = 0, final velocity vf = 26.8 m/s, and t = 10 s, we can find the acceleration using

vf = vi + aDt Þ a =

vf - vi (26.8 m/s - 0 m/s) = = 2.68 m/s2 » 2.7 m/s2 Dt 10 s

(b) The fraction is a/g = 2.68/9.8 = 0.273. So a is 27% of g, or 0.27 g. (c) The displacement is calculated as follows:

1 1 xf - xi = vi Dt + a(Dt)2 = a(Dt)2 = 134 m = 440 feet 2 2 Assess: A little over tenth of a mile displacement in 10 s is physically reasonable.

2-20

Chapter 2

P2.30. Strategize: We will assume acceleration is constant, such that we can use kinematic equations. Prepare: Fleas are amazing jumpers; they can jump several times their body height—something we cannot do. We assume constant acceleration so we can use the kinematic equations. The last of the three relates the three variables we are concerned with in part (a): speed, distance (which we know), and acceleration (which we want). (v y )2f = (v y )2i + 2ay Dy In part (b) we use Equation 2.11 because it relates the initial and final velocities and the acceleration (which we know) with the time interval (which we want).

(v y )f = (v y )i + ay Dt Part (c) is about the phase of the jump after the flea reaches takeoff speed and leaves the ground. So now it is (v y )i , that is 1.0 m/s instead of (v y )f . And the acceleration is not the same as in part (a)—it is now -g (with the positive direction up) since we are ignoring air resistance. We do not know the time it takes the flea to reach maximum height, so we employ Equation 2.13 again because we know everything in that equation except Dy. Solve: (a) Use (v y )i = 0.0 m/s and rearrange Equation 2.13. ay =

(v y )f2 2 y

(1.0 m/s) 2  1000 mm  2   = 1000 m/s 2(0.50 mm)  1 m 

(b) Having learned the acceleration from part (a) we can now rearrange Equation 2.12 to find the time it takes to reach takeoff speed. Again use (vy )i = 0.0 m/s.

Dt =

(v y )f ay

1.0 m/s = .0010 s 1000 m/s 2

Assess: Just over 5 cm is pretty good considering the size of a flea. It is about 10–20 times the size of a typical flea. Check carefully to see that each answer ends up in the appropriate units.

P2.31. Strategize: This is a question about acceleration and how it relates to other kinematic quantities. We will assume the large acceleration is constant, such that we can make use of kinematic equations. Reason: We can use Equation 2.11 to relate acceleration to initial and final speeds. To relate the acceleration and time to distance covered, we can use the initial velocity from part (a) and Equation 2.13. Let us call the initial direction of motion the + x direction, such that the acceleration will be in the − x direction. Solve: The maximum initial speed would require the maximum allowed time to stop. So we assume t = 30 ms. Then

( vx )f = ( vx )i + axt  ( vx )i = ( vx )f − axt = ( 0 m/s ) − ( − (50) (9.8 m/s2 )) (30 10−3 s ) = 14.7 m/s We would report our answer to part (a) as 15 m/s. We have kept an extra digit above for use in part (b). To determine the minimum distance, we again assume that all 30 ms of allowable time are used in the stopping process, and we use the initial velocity from part (a), such that we can write

( vx )f = ( vx )i + 2ax x  x = 2

( v x ) f − ( v x )i 2

2a x

( 0 m/s ) − (14.7 m/s ) = 0.22 m 2 ( − ( 50 ) ( 9.8 m/s 2 ) ) 2

Assess: The maximum initial speed we found is around 33 mph. This means that going from full speed to a full stop in 30 ms could be fatal if the initial speed is greater than around 33 mph. Of course, seatbelts, airbags, and crumple zones in cars are designed to increase the distance over which the humans in the car stop to considerably more than 0.22 m. This way humans can survive head-on collisions starting from even greater speeds.

Motion in One Dimension

2-21

P2.32. Strategize: Let us assume the acceleration is constant over the intervals described. Prepare: We’ll do this in parts, first computing the acceleration after the congestion. Solve: Dv 12.0 m/s - 5.0 m/s 7.0 m/s = = Dt 8.0 s 8.0 s Now use the same acceleration to find the new velocity. a=

 7.0  vf = vi + at = 12.0 m/s +  m/s 2  (16 s) = 26 m/s  8.0  Assess: The answer is a reasonable 58 mph.

P2.33. Strategize: Because the skier slows steadily, her acceleration is a constant during the glide and we can use the kinematic equations. Prepare: We can use Equation 2.13 to determine the unknown acceleration. Solve: Since we know the skier’s initial and final speeds and the width of the patch over which she decelerates, we will use vf2 = vi2 + 2a(xf - xi ) Þa=

vf2 - vi2 (6.0 m/s)2 - (8.0 m/s)2 = = -2.8 m/s2 2(5.0 m) 2(xf - xi )

The magnitude of this acceleration is 2.8 m/s2. Assess: A deceleration of 2.8 m/s2 or 6.3 mph/s is reasonable.

P2.34. Strategize: We will assume constant acceleration of both planes, such that we can use the kinematic equations of motion. Prepare: The kinematic equation that relates velocity, acceleration, and distance is (vx )2f = (vx )2i + 2ax Dx. Solve for Dx.

Dx =

(vx )2f - (vx )2i 2ax

Note that (vx )2i = 0 for both planes. Solve: The accelerations are same, so they cancel.  (vx )f2  2 2   ( (vx )f ) jet ( (2vx )f )prop x jet  2ax  jet = = = = 4   xjet = 4 xprop = 4(1/4 mi) = 1 mi 2 2 xprop  (vx )f2  (vx )f )prop (vx ) f )prop ( (    2ax prop

Assess: It seems reasonable to need a mile for a passenger jet to take off.

P2.35. Strategize: Because the car slows steadily, the deceleration is a constant and we can use the kinematic equations of motion under constant acceleration. Prepare: We look for an equation in which we know all but one variable, and find that we can solve this using Equation 2.13. Solve: Since we know the car’s initial and final speeds and the width of the patch over which she decelerates, we will use vf2 = vi2 + 2a( xf − xi ) a=

vf2 − vi2 (0 m/s) 2 − (90 m/s) 2 = = −37 m/s 2 2( xf − xi ) 2(110 m)

The magnitude of this acceleration is 37 m/s2. Assess: A deceleration of 37 m/s2 is impressive; it is almost 4 gs. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2-22

Chapter 2

P2.36. Strategize: Although this problem does not involve constant acceleration throughout the entire time shown, the acceleration is constant on intervals. Prepare: We recall that displacement is equal to area under the velocity graph between ti and tf, and acceleration is the slope of the velocity-versus-time graph. Solve: (a) Using the equation, xf = xi + area under the velocity-versus-time graph between ti and tf we have, x (at t = 1 s) = x (at t = 0 s) + area between t = 0 s and t = 1 s = 0.0 m + (4 m/s)(1 s) = 4.0 m Reading from the velocity-versus-time graph, vx (at t = 1 s) = 4.0 m/s. Also, ax = slope = v/t = 0 m/s2. (b)

x(at t = 3.0 s) = x(at t = 0 s) + area between t = 0 s and t = 3 s = 0.0 m + 4 m/s  2 s + 2 m/s 1s + (1/2)  2 m/s 1s = 11.0 m Reading from the graph, vx (t = 3 s) = 2 m/s. The acceleration is

ax (t = 3 s) = slope =

vx (at t = 4 s) − vx (at t = 2 s) = −2.0 m/s 2 2s

Assess: Due to the negative slope of the velocity graph between 2 s and 4 s, a negative acceleration was expected.

P2.37. Strategize: Let us assume the acceleration of the car is constant. Prepare: A visual overview of the car’s motion that includes a pictorial representation, a motion diagram, and a list of values is shown below. We label the car’s motion along the x-axis. For the driver’s maximum (constant) deceleration, kinematic equations are applicable. This is a two-part problem. We will first find the car’s displacement during the driver’s reaction time when the car’s deceleration is zero. Then we will find the displacement as the car is brought to rest with maximum deceleration.

Solve: During the reaction time when a0 = 0, we can use

1 a (t - t )2 2 0 1 0 = 0 m + (20 m/s)(0.50 s - 0 s) + 0 m = 10 m

x1 = x0 + v0 (t1 - t0 ) +

During deceleration,

v22 = v12 + 2a1 (x2 - x1 )

0 = (20 m/s)2 + 2(–6.0 m/s2)(x2 – 10 m)  x2 = 43 m

She has 50 m to stop, so she can stop in time. Assess: While driving at 20 m/s or 45 mph, a reaction time of 0.5 s corresponds to a distance of 33 feet or only two lengths of a typical car. Keep a safe distance while driving!

P2.38. Strategize: During the phase of constant acceleration, we can use kinematic equations, and during the subsequent phase of no acceleration, we can again use kinematic equations. But we cannot apply kinematic equations across both phases, since the acceleration changes in between.

Motion in One Dimension

2-23

Prepare: Do this in two parts. First compute the distance traveled during the acceleration phase and what speed it reaches. Then compute the additional distance traveled at that constant speed. Solve: During the acceleration phase, since (vx )i = 0 and xi = 0, xf =

1 1 ax (  t ) 2 = (250 m/s 2 )(20 ms) 2 = 0.05 m = 5.0 cm 2 2

We also compute the speed it attains.

vx = ax Dt = (250 m/s2 )(20 ms) = 5.0 m/s Now the distance traveled at a constant speed of 5.0 m/s .

Dx = vx Dt = (5.0 m/s)(30 ms) = 0.15 m = 15 cm Now add the two distances to get the total.

Dxtotal = 5.0 cm + 15 cm = 20 cm Assess: A 20-cm-long tongue is impressive, but possible.

P2.39. Strategize: We will assume that you achieve the maximum magnitude of acceleration possible, and that this acceleration is constant. Prepare: A visual overview of your car’s motion that includes a pictorial representation, a motion diagram, and a list of values is shown below. We label the car’s motion along the x-axis. For maximum (constant) deceleration of your car, kinematic equations hold. This is a two-part problem. We will first find the car’s displacement during your reaction time when the car’s deceleration is zero. Then we will find the displacement as you bring the car to rest with maximum deceleration.

Solve: (a) To find x2, we first need to determine x1. Using x1 = x0 + v0(t1 – t0), we get x1 = 0 m + (20 m/s) (0.50 s – 0 s) = 10 m. Now, with a1 = 10 m/s2, v2 = 0 and v1 = 20 m/s, we can use

v22 = v12 + 2a1 ( x2 − x1 )  0 m2 /s 2 = (20 m/s) 2 + 2(−10 m/s 2 )( x2 − 10 m)  x2 = 30 m The distance between you and the deer is (x3 – x2) or (35 m – 30 m) = 5 m. (b) Let us find v0 max such that v2 = 0 m/s at x2 = x3 = 35 m. Using the following equation,

v22 − v02 max = 2a1 ( x2 − x1 )  0 m2 /s 2 − v02 max = 2(−10 m/s 2 )(35 m − x1 ) Also, x1 = x0 + v0 max (t1 – t0) = v0 max (0.50 s – 0 s) = (0.50 s)v0 max. Substituting this expression for x1 in the above equation yields

-v02 max = (-20 m/s2 )[35 m - (0.50 s) v0 max ] Þ v02 max + (10 m/s)v0 max - 700 m 2 /s2 = 0 The solution of this quadratic equation yields v0 max = 22 m/s. (The other root is negative and unphysical for the present situation.) Assess: An increase of speed from 20 m/s to 22 m/s is very reasonable for the car to cover an additional distance of 5 m with a reaction time of 0.50 s and a deceleration of 10 m/s 2. © Copyright 2019 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2-24

Chapter 2

P2.40. Strategize: Here we must relate acceleration to other kinematic variables like distance and velocity. Let us assume that the acceleration during the compression of the bag is constant, such that we can use kinematic equations. Prepare: We are given initial and final vertical components of the velocity, as well as the distance of the compression. Equation 2.13 allows us to solve for the acceleration. Let us choose the vertically upward direction to be + y. Solve: Applying Equation 2.13, and paying special attention to the signs of quantities like the compression, we find

( v ) − ( v ) = ( 0 m/s ) − ( −6.0 m/s ) = 150 m/s + 2a y  a = 2

( ) ( ) vy

= vy

2 ( −0.12 m )

2y

We are asked to report this in units of g , so we divide by 9.8 m/s2 to obtain a y = 15 g. Assess: This is much less than the particularly dangerous of 60g given in the text.

P2.41. Strategize: We will assume constant acceleration, such that we can use kinematic equations. Prepare: Call the point where the motorcycle started the origin. Solve: (a) a=

v v 80 km/h  1 h  1000 m   t = =  = 2.78 s  2.8 s t a 8.0 m/s 2  3600 s   1 km 

(b) Compute the distance traveled in 10 s for each vehicle.  1 h  1000 m  For the car: x = vt = (80 km/h)(2.78 s)    = 61.7 m  3600 s  1 km 

(

( )

)(

)

2 2 1 1 a Dt = 8.0 m/s2 2.78 s = 30.7 m 2 2 The difference is the distance between the motorcycle and the car at that time. 61.7 m – 30.7 m = 31 m Assess: The motorcycle will never catch up if it never exceeds the speed of the car.

For the motorcycle: Dx =

P2.42. Strategize: In this problem we will use the slope of a velocity vs. time graph to determine acceleration, and other kinematic variables. The acceleration is assumed to be approximately constant for this insect. Prepare: We can answer part (a) by using Equation 2.11, which is equivalent to determining the slope of a velocity vs. time graph in the case of a constant acceleration. We can then use Equation 2.12 to determine the distance covered by the insect. Solve: (a) The acceleration is given by ay =

( v ) − ( v ) = ( 0.90 m/s ) − ( 0 m/s ) = 1.8 10 m/s t (5.0 10 s ) y

−3

(b) The distance can be determined by

( )

y = v y t + i

2 1 1 2 a y ( t ) = ( 0 m/s ) 5 10−3 m/s + 180 m/s 2 5 10 −3 m/s = 2.3 10 −3 m 2 2

(

)

(

)(

)

Assess: The distance covered is very reasonable for an insect in the process of jumping.

P2.43. Strategize: During the acceleration phase, acceleration is constant, and we can use kinematic equations. Once the acceleration drops to zero, acceleration will once again be constant, and we can apply kinematic equations over the phase of constant velocity. But we cannot apply kinematic equations from the beginning of the dash to the end, since acceleration changes in between. Prepare: Use Equation 2.11 to find the acceleration. vx = axt1 ax =

where v0 = 0 and t0 = 0 vx 11.2 m/s = = 5.23 m/s2 t1 2.14 s

Motion in One Dimension

2-25

Solve: The distance traveled during the acceleration phase will be 1 a (Dt)2 2 x 1 = (5.23 m/s 2 )(2.14 s)2 2 = 12.0 m

Dx =

The distance left to go at constant velocity is 100 m -12.0 m = 88.0 m. The time this takes at the top speed of 11.2 m/s is

Dt =

Dx 88.0 m = = 7.86 s vx 11.2 m/s

The total time is 2.14 s + 7.86 s = 10.0 s. Assess: This is indeed about the time it takes a world-class sprinter to run 100 m (the world record is a bit under 9.8 s). Compare the answer to this problem with the accelerations given in Problem 2.25 for Carl Lewis.

P2.44. Strategize: In this problem we consider hoverflies falling under the influence of gravity, which yields a constant acceleration. Thus we can use kinematic equations. Prepare: The hoverflies are released from rest. We can find the distance they would cover if they fell only under the influence of gravity (no wing flapping) using Equation 2.12 and inserting the specified time of 200 ms. For part (b), we can use the same equation again. Only this time we will insert a known distance we want the hoverflies to cover and solve for the time. 1 1 2 2 Solve: For part (a):  y = v y  t + a y (  t ) = ( 0 m/s )( 0.200 s ) + −9.8 m/s 2 ( 0.200 s ) = −0.20 m. So the i 2 2 hoverflies would have fallen 20 cm in the first 200 ms. 1 1 2 2 For part (b)  y = v y  t + a y (  t )   y = a y (  t ) since the hoverflies are dropped, not thrown with some i 2 2

(

( )

)

( )

initial velocity. Then t =

2 y = ay

2 ( −0.40 m )

( −9.8 m/s ) 2

= 0.29 s or 290 ms.

Assess: The result from (b) explains why only some hoverflies avoided hitting the bottom when began flying after 200 ms, since the entire fall takes only 290 ms.

P2.45. Strategize: This problem involves freefall, in which the acceleration is constant. Thus, we can use the kinematic equations. Prepare: The bill must drop its own length in freefall. Solve: Dy =

1 2Dy 2(0.16 m) g(Dt) 2 Þ Dt = = = 0.18 s 2 g 9.8 m/s 2

Assess: This is less than the typical 0.25 s reaction time, so most people miss the bill.

P2.46. Strategize: This problem involves freefall, in which the acceleration is constant. Thus, we can use the kinematic equations. Prepare: We will assume that, as stated in the chapter, the bill is held at the top, and the other person’s fingers are bracketing the bill at the bottom. Call the initial position of the top of the bill the origin, y0 = 0.0 m, and, for convenience, call the down direction positive. In free fall the acceleration a y will be 9.8 m/s2. The length of the bill will be Dy, the distance the top of the bill can fall from rest in 0.25 s.

2-26

Chapter 2

Solve: yf =

1 1 a y ( t ) 2 = (9.8 m/s 2 )(0.25 s) 2 = 0.31 m 2 2

Assess: This is about twice as long as real bills are (they are really 15.5 cm long), so if a typical reaction time is 0.25 s, then almost no one would catch one in this manner. To catch a bill as small as real bills, one would need a reaction time of 0.13 s.

P2.47. Strategize: This problem involves freefall, in which the acceleration is constant. Thus, we can use the kinematic equations. Prepare: Use kinematic equations for constant acceleration. Assume the gannet is in free fall during the dive. Solve: (v y )f2 = (v y )i2 + 2 g y  y =

(v y )f2 2g

(32 m/s) 2 = 52 m 2(9.8 m/s 2 )

Assess: 52 meters seems a reasonable height from which to begin the dive.

P2.48. Strategize: Once the trout leaves the water, it is subject to acceleration due to gravity only. The acceleration is constant, so we can use kinematic equations. Prepare: In order to find the maximum height, we can use Equation 2.13. We recognize that at the maximum height the vertical component of the velocity is momentarily zero. Let us call the vertically upward direction + y. For part (b), we must consider that the maximum height may be greater than 1.8 m. That means the trout will rise to its maximum height and then fall back to 1.8 m. Solve: (a) To find the maximum height, we write

( ) ( ) vy

= vy

2 i

( v ) − ( v )  y + 2a  y   y = y

y f

y i

2a y

( 0 m/s ) − (8.0 m/s ) = 3.3 m. 2

max =

(

2 −9.8 m/s 2

)

(b) One way of finding the time required to reach a height of 1.8 m is to use Equation 2.12 and solve the quadratic equation for time:

( )

− vy  1 2 i 0 = − y + v y  t + a y (  t )   t = i 2

( )

t =

− ( 8.0 m/s ) 

( v ) + 2ya 2

y i

(8.0 m/s ) + 2 (1.8 m ) ( −9.8 m/s 2 ) 2

( −9.8 m/s ) 2

= 0.27 s or 1.4 s

Clearly the trout rises past a height of 1.8 m before reaching its maximum height and falling back to a height of 1.8 m. Hence, we want the later of the two times that the trout reaches this height: 1.4 s. Assess: This a reasonable time of flight for a fish jumping out of the water.

P2.49. Strategize: Two objects are in freefall. We must determine where they meet. Since the acceleration is constant during freefall, we can use kinematic equations. Prepare: We will need to describe the motion of the acrobat and the ball, for which we will use subscripts A and B, respectively. Let us call the vertically upward direction + y. To determine when the acrobat catches the ball, we must determine when the two objects have the same vertical positions. We can use Equation 2.12 to describe the change in position of each object. In order for them to meet, we require yA = yB + ( 9.0 m ). That is, however much the ball moves, the acrobat must move upward by 15 m more to cover the initial distance between them.

Motion in One Dimension

2-27

Solve: We apply Equation 2.12 to each object separately:

( )

 yA = vA,y  t +

( )

 yB = vB,y  t + i

1 2 a y ( t ) 2

1 1 2 2 a y ( t ) = a y ( t ) 2 2

In the second equation we have use the fact that the ball is dropped, not thrown. Note that no subscript is required for the time or the acceleration, since we need the acrobat and ball to be in the same place at the same time, and since both are accelerating only due to gravity. Requiring yA = yB + ( 9.0 m ) , we find

( v ) t + 12 a ( t ) = ( 9.0 m ) + 12 a ( t ) 2

A, y i

Subtracting

1 2 a y ( t ) from both sides yields 2

(v ) t = (9.0 m)  t = (9.0 m) / (v ) = (9.0 m) / (8.0 m/s ) = 1.1 s. A,y

A,y

Assess: Given the length scales of sever meters, a time of 1.1 s for the two objects to meet is reasonable.

P2.50. Strategize: If we ignore air resistance then the only force acting on both balls after they leave the hand (before they land) is gravity; they are therefore in free fall. Prepare: Think about ball A’s velocity. It decreases until it reaches the top of its trajectory and then increases in the downward direction as it descends. When it gets back to the level of the student’s hand it will have the same speed downward that it had initially going upward; it is therefore now just like ball B (only later). Solve: (a) Because both balls are in free fall they must have the same acceleration, both magnitude and direction, 9.8 m/s2, down. (b) Because ball B has the same downward speed when it gets back to the level of the student that ball A had, they will have the same speed when they hit the ground. Assess: Draw a picture of ball B’s trajectory and draw velocity vector arrows at various points of its path. Air resistance would complicate this problem significantly. P2.51. Strategize: Once the jumper leaves the ground, he or she is in freefall, in which the acceleration is constant. Thus, we can use the kinematic equations. Prepare: Use the kinematic equation (v y )f2 = (v y )i2 + 2a y y where (v y )2f = 0 at the top of the leap. We assume a y = -9.8 m/s2 and we are given Dy = 1.1 m. Solve:

(vy )i2 = −2ay y  (vy )i = −2ay y = −2(−9.8 m/s 2 )(1.1 m) = 4.6 m/s Assess: This is an achievable take-off speed for good jumpers. The units also work out correctly and the two minus signs under the square root make the radicand positive.

P2.52. Strategize: Once the ball is in the air, it is in freefall, in which acceleration is constant. Thus we can use kinematic equations. Prepare: Assume the trajectory is symmetric (i.e., the ball leaves the ground) so half of the total time is the upward portion and half downward. Put the origin at the ground. Assume no air resistance. Solve: (a) On the way down (v y )i = 0 m/s, yf = 0 m, and Dt = 2.6 s. Solve for yi . 0 = yi +

1 a y (t ) 2 2



1 1 yi = − a y ( t ) 2 = − ( − 9.8 m/s 2 )(2.6 s) 2 = 33.1 m 2 2

or 33 m to two significant figures.

2-28

Chapter 2

(b) On the way up (v y )f = 0 m/s.

(vy )i2 = −2ay y  (vy )i = −2ay y = −2(−9.8 m/s2 )(33.1 m) = 25 m/s Assess: When thinking about real football games, this speed seems reasonable.

P2.53. Strategize: Once the briefcase is dropped it is in freefall, during which the acceleration is constant. Thus we can use kinematic equations. Prepare: Since the villain is hanging on to the ladder as the helicopter is ascending, he and the briefcase are moving with the same upward velocity as the helicopter. We can calculate the initial velocity of the briefcase, which is equal to the upward velocity of the helicopter. See the following figure.

Solve: We can use Equation 2.12 here. We know the time it takes the briefcase to fall, its acceleration, and the distance it falls. Solving for (v y )i Dt,

é1 ù 1 (v y )i Dt = ( yf - yi ) - (a y )Dt 2 = -130 m - ê (-9.80 m/s2 )(6.0 s)2 ú = 46 m 2 ë2 û Dividing by Dt to solve for (v y )i ,

(vx )i =

46 m = 7.7 m/s 6.0 s

Assess: Note the placement of negative signs in the calculation. The initial velocity is positive, as expected for a helicopter ascending.

P2.54. Strategize: Assume the jumper is in free fall after leaving the ground, so that we can use the kinematic equations. Prepare: We assume a y = -9.8 m/s2 and we are given ( yf − yi ) = 1.1 m. We can use Equation 2.12 to determine the unknown time. Solve: Since the trajectory is symmetric we’ll compute the time it takes to come down from 1.1 m to the floor and then double it. ( yf − yi ) =

1 a y (t ) 2  t = 2

2( yf − yi ) = ay

2( − 1.1 m) = 0.47 s −9.8 m/s 2

The whole “hang time” will be double this, or 0.95 s. Assess: This is about the time for a big leap. The units also work out correctly and the two minus signs under the square root make the radicand positive.